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Titration Calculator with Steps (Free) | StatsUnlock

Titration Calculator

Calculate concentration from titration data with concordant titres and uncertainty, standardize a titrant, predict the endpoint, plot the full pH titration curve for strong, weak and polyprotic acids and bases with equivalence points and indicators, find the molar mass of an unknown acid, and solve back titrations, double titrations and titratable acidity, with every step, a figure and four charts.

LabChemistryStep by StepAcid–BasepH CurveEquivalence PointA Level and APFree

⚡ 0. Quick Answer

A titration calculator uses moles = concentration × volume and the mole ratio from the balanced equation: c(analyte) = c(titrant) × V(titrant) × a/b ÷ V(analyte). For 25.00 mL of HCl needing a mean concordant titre of 24.55 mL of 0.1000 M NaOH: 0.1000 × 24.55 ÷ 25.00 = 0.09820 mol/L HCl (± 0.00042, k = 2).

n = c × V  ·  caVa ÷ a = ctVt ÷ b  ·  Veq = nanalyte × b/a ÷ ct  ·  pH at half-equivalence ≈ pKa

Rule of thumb: ignore the rough titre, average only concordant titres (within 0.10 mL), convert mL to L, and use the mole ratio: H2SO4 needs twice as much NaOH as HCl.

⭐ Key Takeaways

  • A titration calculator turns titres of 24.80 (rough), 24.55, 24.60 and 24.50 mL into a mean of 24.55 mL and 0.09820 M HCl.
  • For H2SO4 + 2NaOH, divide the moles of NaOH by 2: the ratio changes the answer by a factor of two.
  • 0.1 M acetic acid with 0.1 M NaOH starts at pH 2.88, passes pH 4.76 (= pKa) at half-equivalence and reaches pH 8.73 at equivalence: use phenolphthalein.
  • 0.1500 g of a diprotic acid needing 25.40 mL of 0.1000 M NaOH has a molar mass of 118.1 g/mol.
  • Titratable acidity: 15.60 mL of 0.1 N NaOH for 10 mL of juice is 1.00% citric acid.

📚 1. What Is a Titration and How Do You Calculate It?

A titration measures an unknown amount of a substance (the analyte) by adding a solution of known concentration (the titrant) from a burette until the reaction is just complete: the equivalence point, seen as the endpoint by an indicator color change or a pH jump. Because moles = concentration × volume and the balanced equation fixes the mole ratio, one volume reading gives the unknown concentration. Students often find the mole ratio and the meaning of the equivalence point the hardest parts (Sheppard, 2006). A titration calculator handles the arithmetic, the concordant titres, the uncertainty and, for acid–base titrations, the full pH curve from an exact charge balance (de Levie, 1999).

A worked one-liner: n(NaOH) = 0.1000 × 24.55 ÷ 1000 = 2.455 × 10−3 mol = n(HCl); c(HCl) = 2.455 × 10−3 ÷ 0.02500 = 0.09820 mol/L.

Who uses it: GCSE, A level, IB and AP chemistry students writing up titration practicals; university analytical chemistry labs; quality control in food, pharmaceutical and water labs; and anyone checking an acid–base titration lab report, choosing an indicator or predicting the endpoint before running a titration.

Quick answers to the questions people ask most

How do you calculate concentration from titration?

Find the moles of titrant (concentration × titre in liters), use the balanced equation’s mole ratio to get moles of analyte, then divide by the analyte volume in liters. With 24.55 mL of 0.1000 M NaOH for 25.00 mL of HCl, n = 0.002455 mol and c(HCl) = 0.09820 mol/L.

How do you calculate the average titre?

Leave out the rough (first) titration, then average only concordant titres: those within 0.10 mL of each other. From 24.80 (rough), 24.55, 24.60 and 24.50 mL, the concordant titres are 24.50, 24.55 and 24.60, so the mean titre is 24.55 mL. Never average a titre that disagrees.

How do you calculate the equivalence point of a titration?

The equivalence point volume is the moles of analyte times the mole ratio, divided by the titrant concentration. For 25.0 mL of 0.0800 M HCl and 0.100 M NaOH, V = 0.00200 mol ÷ 0.100 M = 20.0 mL. Calculating it before titrating tells you where to slow down.

How do you calculate pH during a titration?

Before equivalence, a weak acid forms a buffer; at half-equivalence pH equals pKa. At equivalence, the pH depends on the salt formed. After equivalence, excess titrant sets the pH. This calculator solves the full charge balance at every volume, so one method covers all regions exactly.

How do you calculate molar mass from titration?

Moles of acid equal moles of NaOH divided by the number of acidic protons. Divide the mass of acid by those moles. For 0.1500 g of a diprotic acid neutralized by 25.40 mL of 0.1000 M NaOH, moles of acid = 0.001270 and the molar mass is 118.1 g/mol.

How do you calculate concentration of H2SO4 in titration?

Use the 1 : 2 ratio from H2SO4 + 2NaOH → Na2SO4 + 2H2O. Moles of acid are half the moles of NaOH. A mean titre of 22.13 mL of 0.100 M NaOH for 25.0 mL of acid gives 0.002213 mol NaOH, 0.001106 mol H2SO4 and 0.0443 mol/L.

Bench note from the StatsUnlock Team: three slips cost most marks in titration write-ups. Averaging the rough titre with the accurate ones; forgetting to divide mL by 1000; and ignoring the mole ratio for diprotic acids or redox titrations (MnO4− reacts with five Fe2+). In the lab, the burette reading and the endpoint judgment dominate the uncertainty, which the budget in Section 3 makes visible (Schwartz, 1992).
How a titration calculator turns a burette reading into a concentration 24.55 milliliters of 0.1000 molar NaOH contains 0.002455 moles, which reacts one to one with HCl in a 25.00 milliliter aliquot, giving 0.09820 molar. Titrant0.1000 M × 24.55 mLmean concordant titre Moles0.002455 mol NaOH× 1/1 = mol HCl Analyte÷ 0.02500 L0.09820 M HCl n = c × V (in liters) · mole ratio from the balanced equation · c = n ÷ V
Every titration calculation: volume to moles, moles across the equation, moles to concentration.
TitrationBalanced equationAnalyte : titrantIndicator
HCl with NaOHHCl + NaOH → NaCl + H2O1 : 1phenolphthalein or methyl orange
H2SO4 with NaOHH2SO4 + 2NaOH → Na2SO4 + 2H2O1 : 2phenolphthalein
Acetic acid with NaOHCH3COOH + NaOH → CH3COONa + H2O1 : 1phenolphthalein
NH3 with HClNH3 + HCl → NH4Cl1 : 1methyl red
Na2CO3 with HClNa2CO3 + 2HCl → 2NaCl + H2O + CO21 : 2methyl orange
Fe2+ with KMnO4MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O5 : 1self-indicating

🧮 2. Enter Your Titration Data

Optional: molar mass, equivalents and uncertainty

Mole ratio from the balanced equation: for H₂SO₄ + 2NaOH enter a = 1 (acid), b = 2 (NaOH).

📊 3. Results and Step-by-Step Working

Load a sample or enter your titration data above, then press Calculate Titration. The concentration, moles, the formula, every step, a figure, four charts and the exports will appear here.

🧠 4. Understanding Your Titration Result, In Detail

4.1 What the result means

___

4.2 The formula

___

4.3 The detail that changes the answer

___

4.4 Accuracy and uncertainty

___

4.5 What to do next

___

4.6 Endpoint versus equivalence point

The equivalence point is where the moles of titrant exactly match the analyte according to the equation; the endpoint is where you see the change. A well-chosen indicator changes color on the steep part of the curve, so the two differ by less than a drop. For weak endpoints, a pH meter and a first-derivative or Gran plot locate the equivalence point more precisely (Gran, 1952; Rossotti & Rossotti, 1965).

4.7 Why exact pH curves

Textbook pH calculations split the curve into regions (strong acid, buffer, salt, excess base) with a different approximation for each. Solving the full charge balance at every volume gives one exact expression that works for strong, weak and polyprotic systems and mixtures (de Levie, 1993; Asuero & Michałowski, 2011). It also shows when an equivalence point is too flat to detect, as for the third proton of phosphoric acid.

4.8 Choosing the titrant

Titrating a weak base with a weak acid gives no sharp endpoint, so one partner is always strong. The choice of titrant and receiver solution still matters in methods such as Kjeldahl nitrogen, where the theory of the titration curve guides it (Michałowski et al., 2013).

✍ 5. How to Write Titration Results in a Lab Report: Detailed Results

A titration result in a report needs the concentration with units, the mean concordant titre and how many titres it came from, the titrant concentration, the indicator or endpoint method, and the uncertainty. The seven write-ups below are filled from the titration calculator; they show the headline example until you press Calculate.

Showing the headline example (HCl with 0.1000 M NaOH). Press Calculate above to fill all seven write-ups with your own result.

5.1 Style rules

RuleWriteAvoid
Record burette readings to 0.05 mL24.55 mL24.5 mL, 24.553 mL
State which titres were averagedmean of 3 concordant titresaverage titre
Give units and concentration basis0.09820 mol/L (mol dm⁻³)0.0982
Show the mole ratio1 : 2 (H₂SO₄ : NaOH)no equation
Significant figures follow the data4 s.f. from 4 s.f. volumes0.098200000
Name the endpoint methodphenolphthalein, colorless to faint pink“the endpoint”

5.2 Seven ready-to-copy write-ups

1. Lab report or paper sentence
The analyte concentration was 0.09820 ± 0.00042 mol/L (U, k = 2), from a mean concordant titre of 24.55 mL (n = 3) with a 25.00 mL aliquot and a 1:1 mole ratio.
2. Data sheet or log entry
Concentration of the analyte from titration Analyte concentration: 0.09820 mol/L Mean titre: 24.55 mL Titres used: 24.50, 24.55, 24.60 mL Moles of titrant: 0.002455 mol Moles of analyte: 0.002455 mol Mole ratio: 1 : 1 (analyte : titrant) Mass concentration: 3.580 g/L SD of titres: 0.0500 mL Check: c(titrant) V(titrant) × a/b = 0.0024550 mol = c(analyte) V(analyte) ✓
3. Code and spreadsheet formulas
Python: c_a = 0.1000 * 24.55 * 1 / 1 / 25.00 R: c_a <- 0.1000 * mean(titres) * 1 / 1 / 25.00 Excel: =A2*AVERAGE(C2:E2)*1/1/B2 JavaScript: const c = 0.1000 * 24.55 * 1 / 1 / 25.00;
4. Exam or homework answer, fully worked
Given: c(titrant) 0.1000 M, V(analyte) 25.00 mL, titres 24.80, 24.55, 24.60, 24.50 mL Required: Analyte concentration Equation: c(a) = c(t) V(t) × a/b ÷ V(a) Substitute: 0.1000 × 24.55 × 1/1 ÷ 25.00 Solve: 0.09820 ± 0.00042 mol/L Working: 1. Rough titre excluded: 24.80 mL (rough or trial run) 2. Concordant titres: 24.50, 24.55, 24.60 mL (range 0.10 mL ≤ 0.10) 3. Mean titre: (24.50 + 24.55 + 24.60) ÷ 3 = 24.55 mL 4. Moles of titrant: 0.1000 × 24.55 ÷ 1000 = 0.002455 mol 5. Moles of analyte: 0.002455 × 1/1 = 0.002455 mol 6. Analyte concentration: 0.002455 ÷ (25.00 ÷ 1000) = 0.09820 mol/L 7. Mass concentration: 0.09820 × 36.46 = 3.580 g/L 8. Uncertainty: u_rel = √(Σ (u/x)²) = 0.214%; U (k = 2) = ± 0.00042 mol/L 9. Check: c(titrant) V(titrant) × a/b = 0.0024550 mol = c(analyte) V(analyte) ✓ Answer: 0.09820 ± 0.00042 mol/L
5. Plain text for chat or a forum
Concentration of the analyte from titration = 0.09820 ± 0.00042 mol/L (Mole ratio 1 : 1 (analyte : titrant)).
6. Detailed results paragraph
Results. Concentration of the analyte from titration = 0.09820 ± 0.00042 mol/L. Mean titre 24.55 mL (3 concordant) · analyte 0.002455 mol · 3.580 g/L. Supporting values: Analyte concentration: 0.09820 mol/L; Mean titre: 24.55 mL; Titres used: 24.50, 24.55, 24.60 mL; Moles of titrant: 0.002455 mol; Moles of analyte: 0.002455 mol; Mole ratio: 1 : 1 (analyte : titrant); Mass concentration: 3.580 g/L; SD of titres: 0.0500 mL. Check: c(titrant) V(titrant) × a/b = 0.0024550 mol = c(analyte) V(analyte) ✓.
7. Results table (tab-separated, pastes into Excel or Word)
Quantity Value Note Analyte concentration 0.09820 mol/L ± 0.00042 (U, k = 2) Mean titre 24.55 mL 3 concordant Titres used 24.50, 24.55, 24.60 mL rough 24.80 excluded Moles of titrant 0.002455 mol Moles of analyte 0.002455 mol Mole ratio 1 : 1 (analyte : titrant) Mass concentration 3.580 g/L M = 36.46 g/mol SD of titres 0.0500 mL

∑ 6. What Is the Acid–Base Titration Calculation Formula? All Formulas Used

①
Concentration from titration
nt = ctVt  ·  na = nt × a/b  ·  ca = na ÷ Va (volumes in L)
a, bcoefficients of analyte and titrant in the balanced equation
Example0.1000 × 0.02455 × 1/1 ÷ 0.02500 = 0.09820 M
②
Mean titre and concordance
mean = ΣV ÷ n over titres agreeing within 0.10 mL; rough titre excluded
Example(24.50 + 24.55 + 24.60) ÷ 3 = 24.55 mL
③
Equivalence point and expected endpoint
Veq = caVa × b/a ÷ ct  ·  polyprotic: Veq,k = k × na ÷ ct
Example0.0800 × 25.0 ÷ 0.100 = 20.0 mL
④
pH during the titration (exact)
[H+] + [Na+] + … = Kw/[H+] + CA × (mean protons lost) + [Cl−] + … solved for [H+]
Regionsinitial; buffer (pH ≈ pKa + log[A⁻]/[HA]); half-equivalence pH ≈ pKa; equivalence; excess titrant
Example0.1 M acetic acid: pH 2.88 → 4.76 → 8.73 at equivalence
⑤
Molar mass of an unknown acid
n(acid) = cNaOHVNaOH ÷ protons × Vflask/Valiquot  ·  M = mass ÷ n(acid)
Example0.1500 g ÷ (0.002540 ÷ 2) = 118.1 g/mol
⑥
Back titration and double titration
n(reacted) = c1V1 − c2V2  ·  Warder: V1 > V2: n(CO₃²⁻) = cV2, n(OH⁻) = c(V1 − V2)
Example50.00 mmol HCl − 20.00 mmol excess = 30.00 mmol; CaCO₃ 15.00 mmol = 1.501 g
⑦
Titratable acidity and uncertainty
% acid = V × N × meq weight ÷ sample × 100  ·  urel(c) = √Σ(ui/xi)², U = 2u
Example15.60 × 0.1000 × 0.06404 ÷ 10.0 × 100 = 1.00% citric acid

🔀 7. How Do You Calculate a Titration? Flowchart, Step by Step

The same worked problem as Sections 0, 9, 10, 11 and 14: 25.00 mL HCl, 0.1000 M NaOH, titres 24.80 (rough), 24.55, 24.60, 24.50 mL.

Start or resultData stepCalculation stepYes or no check
StartTitration results24.80 (rough), 24.55, 24.60, 24.50 mL
CheckDo at least two titres agree within 0.10 mL?
Yes24.50, 24.55, 24.60 (range 0.10).
NoTitrate again.
1
Mean titre(24.50 + 24.55 + 24.60) ÷ 324.55 mL
2
Moles of NaOH0.1000 × 24.55 ÷ 10002.455 × 10⁻³ mol
3
Moles of HCl (1 : 1)× 1/12.455 × 10⁻³ mol
4
Concentration2.455 × 10⁻³ ÷ 0.02500 = 0.09820 mol/L
CheckDoes caVa equal ctVt?
Yes0.09820 × 25.00 = 0.1000 × 24.55. Verified.
NoCheck the ratio and mL to L.
Resultc(HCl) = 0.09820 ± 0.00042 mol/L3.580 g/L; U at k = 2

📝 8. How to Use This Titration Calculator

  1. Pick what you want to calculate. Choose a mode in Section 2: concentration, pH titration curve, molar mass of an unknown acid, back titration, double titration or titratable acidity.
  2. Load a sample or type your data. Twelve worked problems are in the sample menu. Sample 1 is the headline HCl and NaOH titration. Nothing is calculated until you press Calculate Titration.
  3. Say what you are solving for. In the concentration mode, find the analyte concentration, standardize the titrant, or predict the endpoint volume before titrating.
  4. Enter the mole ratio. Type the coefficients of the analyte (a) and titrant (b) from the balanced equation: 1 and 2 for H2SO4 with NaOH, 5 and 1 for Fe2+ with KMnO4.
  5. Enter all titres in order. List every titre separated by commas. Choose whether the first is a rough run; the calculator keeps the largest concordant set within 0.10 mL.
  6. Add tolerances for the uncertainty. Optional: burette, pipette and titrant tolerances give an expanded uncertainty and a budget showing the biggest source.
  7. Press Calculate Titration. Press the button or Enter. The concentration, moles, a Verified badge, the formula and every step appear in Section 3.
  8. Read the curve and charts. The figure draws the burette or the pH curve; four charts show the titres, the derivative, moles and sensitivity.
  9. Copy the write-up. Section 5 fills seven ready texts for a lab report, a log, code, an exam answer, chat, a results paragraph and a table.
  10. Download or run a batch. Download Doc or PDF, or calculate many titrations at once from a CSV of titrant molarity, titre, aliquot and ratio in the Batch tab.

📈 9. How to Do Titration Calculations in Excel

A titration sheet needs one formula for the mean titre and one for the concentration. The sheet repeats the headline problem, so the numbers match Sections 0, 7, 10, 11 and 14.

The whole thing in one cell: with NaOH molarity in A2, the HCl aliquot in B2 and the three concordant titres in C2:E2, type =A2*AVERAGE(C2:E2)/B2. It returns 0.0982.

9.1 The formulas you need

You wantExcel formulaNote
Mean titre=AVERAGE(C2:E2)Concordant titres only
Concordance check=MAX(C2:E2)-MIN(C2:E2)<=0.1TRUE if concordant
Moles of titrant=A2*AVERAGE(C2:E2)/1000mol
Analyte concentration=A2*AVERAGE(C2:E2)*a/b/B2replace a/b
Expected endpoint=G2*B2*b/a/A2G2 = estimated analyte M
Molar mass of acid=H2/(A2*I2/1000/J2)mass, titre, protons
Back titration moles=(K2*L2-M2*N2)/1000reagent minus excess
Titratable acidity=O2*P2*0.06404/Q2*100citric acid
Relative uncertainty=SQRT((0.05/SQRT(3)/F2)^2+(0.03/SQRT(3)/B2)^2)add more terms

9.2 Step by step

1 Enter the data

NaOH molarity in A2, HCl aliquot in B2.

titration.xlsxB2fx25.00AB1c NaOHV HCl20.100025.00

A2 0.1000, B2 25.00.

2 Enter the concordant titres

Rough titre left out: C2 to E2.

titration.xlsxE2fx24.50ABCDE1c NaOHV HClT1T2T320.100025.0024.5524.6024.50

C2 to E2 24.55, 24.60, 24.50.

3 Concentration

In F2 type =A2*AVERAGE(C2:E2)/B2.

titration.xlsxF2fx=A2*AVERAGE(C2:E2)/B2ABCDEF1c NaOHV HClT1T2T3c HCl20.100025.0024.5524.6024.500.09820

F2 shows 0.09820 mol/L.

9.3 Many groups or samples

Put one titration per row and fill the formula down. Add a column with the concordance check so non-concordant rows stand out.

9.4 Why Excel may differ from R or Python

SituationWhat happensFix
Rounded mean titreRounding 24.55 before multiplying changes the last digitRound only the final concentration
STDEV vs STDEV.PRepeatability needs the sample SDUse STDEV.S
pH curvesExcel has no root finder in a cellUse Goal Seek or the calculator

9.5 Charting a pH curve

Enter volume and measured pH in two columns and insert a Scatter chart with smooth lines. Add a column of ΔpH/ΔV to plot the first derivative; its peak marks the equivalence point.

9.6 Errors you may see

  • #DIV/0!: aliquot cell empty.
  • An answer 1000 times too large: mL divided by L, or vice versa.
  • A result double or half the expected: mole ratio missing.

📈 10. How to Do Titration Calculations in R

A complete, copy-and-run script that reproduces the titration calculator for the headline problem. It excludes the rough titre, averages concordant titres, calculates the HCl concentration with its expanded uncertainty, solves the exact pH curve for acetic acid with NaOH, finds the molar mass of an unknown acid and saves one 300 dpi figure.

No packages needed. Base R covers every step, including uniroot() for the pH.

10.1 The complete script

R, base only, saves titration_curve.png

# Titration calculator in base R (StatsUnlock)
concordant <- function(v, rough = TRUE, tol = 0.10) {
  if (rough) v <- v[-1]                           # drop the rough (first) titre
  s <- sort(v); best <- s[1]
  for (i in seq_along(s)) for (j in i:length(s))
    if (s[j] - s[i] <= tol + 1e-9 && (j - i + 1) > length(best)) best <- s[i:j]
  best
}
titres <- c(24.80, 24.55, 24.60, 24.50)            # mL; first is the rough run
used <- concordant(titres); vt <- mean(used)
ct <- 0.1000; va <- 25.00; ratio <- 1             # NaOH M, HCl aliquot mL, a/b
ca <- ct * vt / 1000 * ratio / (va / 1000)
cat(sprintf("concordant: %s; mean %.2f mL; c(HCl) = %.5f mol/L\n", paste(used, collapse = ", "), vt, ca))
# uncertainty: burette 0.05, repeatability, pipette 0.03, NaOH 0.0002 (rectangular)
urel <- sqrt((0.05 / sqrt(3) / vt)^2 + (sd(used) / sqrt(length(used)) / vt)^2 +
             (0.03 / sqrt(3) / va)^2 + (0.0002 / sqrt(3) / ct)^2)
cat(sprintf("U (k = 2) = %.5f mol/L\n", 2 * urel * ca))
# exact pH: 25 mL 0.1 M acetic acid (pKa 4.76) with 0.1 M NaOH
ph_at <- function(v, va = 25, ca = 0.1, cb = 0.1, Ka = 10^-4.76, Kw = 1e-14) {
  V <- va + v; CA <- ca * va / V; Na <- cb * v / V
  f <- function(lh) { h <- 10^lh; h + Na - Kw / h - CA * Ka / (Ka + h) }
  -uniroot(f, c(-16, 2), tol = 1e-12)$root
}
cat(sprintf("pH: start %.2f, half-equivalence %.2f, equivalence %.2f\n", ph_at(0), ph_at(12.5), ph_at(25)))
# molar mass: 0.1500 g diprotic acid, 25.40 mL of 0.1000 M NaOH
cat(sprintf("molar mass = %.1f g/mol\n", 0.1500 / (0.1000 * 25.40 / 1000 / 2)))
png("titration_curve.png", width = 7, height = 4, units = "in", res = 300)
v <- seq(0, 40, 0.2)
plot(v, sapply(v, ph_at), type = "l", lwd = 2, xlab = "NaOH added, mL", ylab = "pH", ylim = c(0, 14))
abline(v = 25, lty = 2); points(25, ph_at(25), pch = 19)
invisible(dev.off())

10.2 Output

concordant: 24.5, 24.55, 24.6; mean 24.55 mL; c(HCl) = 0.09820 mol/L
U (k = 2) = 0.00042 mol/L
pH: start 2.88, half-equivalence 4.76, equivalence 8.73
molar mass = 118.1 g/mol

10.3 Traps

  • sd() is the sample SD, which is right for repeatability.
  • uniroot() needs a bracket where the charge balance changes sign; −16 to 2 in log[H⁺] covers pH −2 to 16.
  • Keep volumes in mL in the formula but divide by 1000 for moles.

10.4 What each part does

BlockPurpose
concordant()Drops the rough titre and keeps the largest set within 0.10 mL
Concentration and Uc = c(t) V(t) × a/b ÷ V(a); relative uncertainties combined in quadrature
ph_at()Charge balance solved for [H⁺] by root finding at each volume
Middle blocksGUM uncertainty and the exact pH at start, half-equivalence and equivalence
Last blockTitration curve of acetic acid with NaOH with the equivalence point marked, saved at 300 dpi

10.5 Adapting it

Change the titres, concentrations and pKa for your titration; ph_at() works for any monoprotic acid and can be extended to polyprotic systems.

📈 11. How to Do Titration Calculations in Python

A complete, copy-and-run script that reproduces the titration calculator for the headline problem. It excludes the rough titre, averages concordant titres, calculates the HCl concentration with its expanded uncertainty, solves the exact pH curve for acetic acid with NaOH, finds the molar mass of an unknown acid and saves one 300 dpi figure.

numpyscipymatplotlib

Install once: pip install numpy scipy matplotlib

11.1 The complete script

Python with NumPy, SciPy and Matplotlib, saves titration_curve.png

# Titration calculator in Python (StatsUnlock)
import numpy as np
from scipy.optimize import brentq
import matplotlib.pyplot as plt

def concordant(v, rough=True, tol=0.10):
    """Drop the rough titre, return the largest set of titres within tol mL."""
    v = sorted(v[1:] if rough else v); best = v[:1]
    for i in range(len(v)):
        for j in range(i, len(v)):
            if v[j] - v[i] <= tol + 1e-9 and j - i + 1 > len(best):
                best = v[i:j + 1]
    return best

used = concordant([24.80, 24.55, 24.60, 24.50])      # mL, first is rough
vt = np.mean(used); ct, va, ratio = 0.1000, 25.00, 1.0
ca = ct * vt / 1000 * ratio / (va / 1000)
print(f"concordant {used}; mean {vt:.2f} mL; c(HCl) = {ca:.5f} mol/L")

urel = np.sqrt((0.05 / np.sqrt(3) / vt) ** 2 + (np.std(used, ddof=1) / np.sqrt(len(used)) / vt) ** 2
               + (0.03 / np.sqrt(3) / va) ** 2 + (0.0002 / np.sqrt(3) / ct) ** 2)
print(f"U (k = 2) = {2 * urel * ca:.5f} mol/L")

def ph_at(v, va=25, ca=0.1, cb=0.1, Ka=10 ** -4.76, Kw=1e-14):
    """Exact pH of a weak acid titrated with NaOH (charge balance)."""
    V = va + v; CA = ca * va / V; Na = cb * v / V
    f = lambda lh: 10 ** lh + Na - Kw / 10 ** lh - CA * Ka / (Ka + 10 ** lh)
    return -brentq(f, -16, 2, xtol=1e-12)

print(f"pH: start {ph_at(0):.2f}, half-equivalence {ph_at(12.5):.2f}, equivalence {ph_at(25):.2f}")
print(f"molar mass = {0.1500 / (0.1000 * 25.40 / 1000 / 2):.1f} g/mol")

v = np.linspace(0, 40, 201)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(v, [ph_at(x) for x in v], lw=2); ax.axvline(25, ls="--"); ax.plot(25, ph_at(25), "o")
ax.set_xlabel("NaOH added, mL"); ax.set_ylabel("pH"); ax.set_ylim(0, 14)
fig.tight_layout(); fig.savefig("titration_curve.png", dpi=300)

11.2 Output

concordant [24.5, 24.55, 24.6]; mean 24.55 mL; c(HCl) = 0.09820 mol/L
U (k = 2) = 0.00042 mol/L
pH: start 2.88, half-equivalence 4.76, equivalence 8.73
molar mass = 118.1 g/mol

11.3 Traps

  • np.std needs ddof=1 for the sample SD.
  • brentq() needs opposite signs at the two ends of the bracket.
  • Solve in log[H⁺], not [H⁺], so the root finder covers 16 orders of magnitude.

11.4 What each part does

BlockPurpose
concordant()Drops the rough titre and keeps the largest set within 0.10 mL
Concentration and Uc = c(t) V(t) × a/b ÷ V(a); relative uncertainties combined in quadrature
ph_at()Charge balance solved for [H⁺] by root finding at each volume
Middle blocksGUM uncertainty and the exact pH at start, half-equivalence and equivalence
Last blockTitration curve of acetic acid with NaOH with the equivalence point marked, saved at 300 dpi

11.5 Adapting it

Change the titres, concentrations and pKa for your titration; ph_at() works for any monoprotic acid and can be extended to polyprotic systems.

📋 12. Titration Reference Tables: Indicators, pKa, Glassware and Curves

12.1 Acid–base indicators

IndicatorpH rangeColor changeUse for
Methyl orange3.1 to 4.4red to yellowstrong acid with weak or strong base
Bromocresol green3.8 to 5.4yellow to blueweak base with strong acid
Methyl red4.4 to 6.2red to yellowweak base with strong acid (NH₃)
Bromothymol blue6.0 to 7.6yellow to bluestrong acid with strong base
Phenol red6.8 to 8.4yellow to redstrong acid with strong base
Phenolphthalein8.2 to 10.0colorless to pinkweak acid with strong base
Thymolphthalein9.3 to 10.5colorless to bluevery weak acid with strong base

Transition ranges are approximate and vary slightly between sources. Last checked: October 2026.

12.2 pKa values used by the calculator (25 °C)

AcidpKa
Hydrochloric acidstrong
Sulfuric acidstrong, 1.99
Acetic acid4.76
Formic acid3.75
Lactic acid3.86
Benzoic acid4.20
Phthalic acid (KHP)2.95, 5.41
Oxalic acid1.25, 4.27
Citric acid3.13, 4.76, 6.40
Phosphoric acid2.15, 7.20, 12.35
Carbonic acid6.35, 10.33
Ammonium (NH₄⁺)9.25 (pKb of NH₃ 4.75)

Typical literature values at 25 °C and low ionic strength; see Goldberg et al. (2002) for buffer acids. Enter your own values in custom mode. Last checked: October 2026.

12.3 Concentration from titre (0.1000 M titrant, 25.00 mL aliquot)

Titre (mL)1 : 1 (HCl, mol/L)1 : 2 (H₂SO₄, mol/L)2 : 1 (analyte : titrant)
10.000.040000.020000.08000
12.500.050000.025000.10000
15.000.060000.030000.12000
17.500.070000.035000.14000
20.000.080000.040000.16000
22.500.090000.045000.18000
24.000.096000.048000.19200
24.550.098200.049100.19640
25.000.100000.050000.20000
27.500.110000.055000.22000
30.000.120000.060000.24000
35.000.140000.070000.28000
40.000.160000.080000.32000

c = 0.1000 × titre × a/b ÷ 25.00. Exact. Last checked: October 2026.

12.4 Titratable acidity factors

AcidMolar mass (g/mol)Acidic Hmeq weight (g/mEq)Typical use
Citric acid, anhydrous192.1230.06404citrus juices
Citric acid monohydrate210.1430.07005
Malic acid134.0920.06704apples, stone fruit
Tartaric acid150.0920.07504grapes, wine
Lactic acid90.0810.09008milk, yogurt
Acetic acid60.0510.06005vinegar
Oleic acid282.4610.28246free fatty acids in oils

meq weight = molar mass ÷ acidic protons ÷ 1000 (Tyl & Sadler, 2017). Last checked: October 2026.

12.5 Class A glassware tolerances

ItemVolumeTolerance (±)Standard u (a/√3)
Burette50 mL0.05 mL0.029 mL
Burette25 mL0.03 mL0.017 mL
Volumetric pipette10 mL0.02 mL0.012 mL
Volumetric pipette20 mL0.03 mL0.017 mL
Volumetric pipette25 mL0.03 mL0.017 mL
Volumetric pipette50 mL0.05 mL0.029 mL

Typical Class A tolerances for burettes (ISO 385) and one-mark pipettes (ISO 648); check the marking on your glassware. Last checked: October 2026.

12.6 pH during the titration of 25 mL of 0.1 M acetic acid with 0.1 M NaOH

NaOH added (mL)pH (exact)Point
02.88start
54.16
12.54.76half-equivalence
205.36
246.14
24.97.16
258.73equivalence
25.110.30
2611.29
3011.96
4012.36

Exact charge balance, pKa 4.76, Kw 1.0 × 10⁻¹⁴; the jump from 24.9 to 25.1 mL is about 3 pH units. Last checked: October 2026.

📈 13. Titration Calculation Examples: Questions and Answers

Eight worked problems that cover what a titration calculator meets: a carbonate standardization, vinegar, a redox titration, two pH curves, an unknown acid, an antacid back titration and the classic mistake. Each number was checked in Python.

1
A LEVEL

Standardize HCl with sodium carbonate

Na₂CO₃ + 2HCl.

25.0 mL of 0.0500 M Na₂CO₃ needs a mean titre of 23.80 mL of HCl (methyl orange).

0.0500 M25.0 mL1.25 mmol× 20.1050 M
QuantityValueNote
Moles of Na₂CO₃1.250 × 10⁻³ mol0.0500 × 0.0250
Moles of HCl2.500 × 10⁻³ mol× 2
c(HCl)0.1050 mol/L÷ 0.02380 L
Moles in the reaction1.250 mmolNa2CO32.500 mmolHClmapping: bar height = mmol
Two HCl per carbonate.

What it means: Without the ratio the answer would be half.

How to write it: "c(HCl) = 0.1050 mol/L."

2
FOOD LAB

Acetic acid in vinegar

Dilute, then titrate.

Vinegar is diluted 10.00 mL to 100.0 mL; a 25.00 mL aliquot needs 18.75 mL of 0.1000 M NaOH.

18.75 mL0.1000 M0.07500 M aliquot× 10 dilution0.7500 M
QuantityValueNote
Diluted solution0.07500 mol/L0.1000 × 18.75 ÷ 25.00
Original vinegar0.7500 mol/L× 100.0/10.00
Acetic acid45.0 g/L = 4.50% w/v× 60.05 g/mol
Concentration0.07500 Mdiluted0.7500 Moriginalmapping: bar height = mol/L
Always undo the dilution.

What it means: Table vinegar is usually about 4 to 8% acetic acid.

How to write it: "Vinegar contains 0.7500 M (4.50% w/v) acetic acid."

3
REDOX

Iron(II) with permanganate

MnO₄⁻ + 5Fe²⁺.

25.00 mL of an Fe²⁺ solution needs 22.40 mL of 0.0200 M KMnO₄ in acid.

0.0200 M22.40 mL0.448 mmol MnO₄⁻× 50.0896 M Fe²⁺
QuantityValueNote
Moles of MnO₄⁻4.480 × 10⁻⁴ mol
Moles of Fe²⁺2.240 × 10⁻³ mol× 5
c(Fe²⁺)0.0896 mol/L÷ 0.02500 L
Moles0.4480 mmolMnO4-2.240 mmolFe2+mapping: bar height = mmol
Ratio a/b = 5/1.

What it means: Permanganate is its own indicator: first permanent pink.

How to write it: "c(Fe²⁺) = 0.0896 mol/L."

4
pH CURVE

Acetic acid with NaOH

Weak acid, strong base.

25 mL of 0.1 M acetic acid is titrated with 0.1 M NaOH.

start pH 2.88half-eq 4.76eq pH 8.7325.0 mLphenolphthalein
QuantityValueNote
Initial pH2.88weak acid
Half-equivalence4.76≈ pKa
Equivalence8.73 at 25.0 mLacetate is basic
pH at key points2.880 start4.760 half-eq8.730 equivalencemapping: bar height = pH
pH at equivalence is above 7.

What it means: Methyl orange would change far too early.

How to write it: "Equivalence at 25.0 mL, pH 8.73."

5
WEAK BASE

Ammonia with HCl

Equivalence below 7.

25 mL of 0.1 M NH₃ is titrated with 0.1 M HCl.

pH 11.12 starthalf-eq 9.25eq pH 5.2825.0 mLmethyl red
QuantityValueNote
Initial pH11.12
Half-equivalence9.25= pKa of NH₄⁺
Equivalence5.28NH₄⁺ is acidic
pH at key points11.12 start9.250 half-eq5.280 equivalencemapping: bar height = pH
Phenolphthalein would fail here.

What it means: Choose methyl red (4.4 to 6.2).

How to write it: "Equivalence at 25.0 mL, pH 5.28."

6
UNKNOWN ACID

Identify a diprotic acid

Molar mass from titration.

0.1500 g of a diprotic acid needs 25.40 mL of 0.1000 M NaOH.

2.540 mmol NaOH÷ 21.270 mmol acid0.1500 g118.1 g/mol
QuantityValueNote
Moles of acid1.270 × 10⁻³ mol2.540 ÷ 2
Molar mass118.1 g/mol0.1500 ÷ 0.001270
Candidatesuccinic acid, 118.090.02% error
Molar mass by proton count59.06 g/molmono118.1 g/moldi177.2 g/moltrimapping: bar height = g/mol
Proton count changes everything.

What it means: Count the equivalence points on a pH curve first.

How to write it: "Molar mass 118.1 g/mol (succinic acid)."

7
BACK TITRATION

Calcium carbonate in an antacid

Excess HCl, back with NaOH.

A 1.600 g tablet reacts with 50.00 mL of 1.000 M HCl; the excess needs 40.00 mL of 0.5000 M NaOH.

50.00 mmol HCl20.00 mmol excess30.00 mmol reacted15.00 mmol CaCO₃93.8%
QuantityValueNote
HCl reacted30.00 mmol50.00 − 20.00
CaCO₃1.501 g15.00 mmol × 100.09
Purity93.8%÷ 1.600 g
HCl50.00 mmoladded20.00 mmolexcess30.00 mmolreactedmapping: bar height = mmol
The difference is what reacted.

What it means: Acid-neutralizing capacity: 30.0 mEq per tablet.

How to write it: "The tablet contains 1.501 g CaCO₃ (93.8%)."

8
COMMON MISTAKE

Including the rough titre

Average concordant titres only.

A student averages all four titres, 24.80, 24.55, 24.60 and 24.50 mL.

mean of 4 = 24.61250.09845 M (wrong)mean of 3 = 24.550.09820 M (right)+0.25%
QuantityValueNote
Wrong0.09845 mol/Lrough included
Right0.09820 mol/Lconcordant only
Error+0.25%bias, not random
Mean titre24.61 mLall four (wrong)24.55 mLconcordant (right)mapping: fuchsia bar = wrong method
The rough titre overshoots.

What it means: Exam mark schemes reject averages that include the rough titre.

How to write it: "c(HCl) = 0.09820 mol/L (mean of 3 concordant titres)."

✏️ 14. How Do You Calculate a Titration by Hand? Steps and Lab Procedure

The written method, calculator keys and the titration procedure, for the headline problem: 25.00 mL HCl, 0.1000 M NaOH, titres 24.80 (rough), 24.55, 24.60, 24.50 mL.

14.1 The written method

  1. Equation. HCl + NaOH → NaCl + H2O (1 : 1).
  2. Mean titre. Exclude 24.80; (24.55 + 24.60 + 24.50) ÷ 3 = 24.55 mL.
  3. Moles of NaOH. 0.1000 × 0.02455 = 2.455 × 10⁻³ mol.
  4. Moles of HCl. 2.455 × 10⁻³ mol (1 : 1).
  5. Concentration. 2.455 × 10⁻³ ÷ 0.02500 = 0.09820 mol/L.
  6. Check. 0.09820 × 25.00 = 0.1000 × 24.55 ✓

14.2 On a calculator

StepKeysDisplay
Mean titre(24.55+24.60+24.50)÷3=24.55
Concentration0.1×Ans÷25=0.0982
Mass concentrationAns×36.46=3.580

14.3 Lab SOP: acid–base titration

Purpose

Determine the concentration of a hydrochloric acid solution by titration with standardized sodium hydroxide.

Equipment and reagents

50 mL Class A burette; 25 mL volumetric pipette and filler; 250 mL conical flasks; white tile; standardized 0.1000 M NaOH; phenolphthalein; distilled water.

Hazards and controls

NaOH and HCl are corrosive and irritant: wear safety goggles and gloves; never pipette by mouth; rinse spills with plenty of water.

Procedure

  1. Rinse the burette with NaOH, fill it, remove air from the tip and record the initial reading to 0.05 mL.
  2. Pipette 25.00 mL of HCl into a conical flask; add 2 to 3 drops of phenolphthalein.
  3. Run a rough titration to find the approximate endpoint.
  4. Repeat accurately, adding dropwise within 1 mL of the endpoint and swirling, until the first permanent faint pink.
  5. Repeat until at least two titres agree within 0.10 mL.
  6. c(HCl) = 0.1000 × mean titre ÷ 25.00.

Record sheet

RunFinal (mL)Initial (mL)Titre (mL)
Rough24.800.0024.80
125.100.5524.55 ✓
224.700.1024.60 ✓
324.850.3524.50 ✓

Acceptance criteria

At least two titres within 0.10 mL; endpoint color stable for about 30 s. These are teaching-lab criteria; follow your course or laboratory method where they differ.

Sources of error and their direction

  • Overshooting the endpoint: titre high, result high.
  • Air bubble in the burette tip: titre high.
  • Burette not rinsed with titrant: titrant diluted, titre high.
  • Pipette not rinsed with analyte: analyte diluted, titre low.
  • Parallax: either way.

14.4 Practice problems

1. 20.00 mL of NaOH needs 18.40 mL of 0.1050 M HCl. c(NaOH)?

0.0966 mol/L

2. 25.0 mL H2SO4 needs 20.0 mL of 0.150 M NaOH. c(H2SO4)?

0.0600 mol/L

3. Titres 23.10 (rough), 22.65, 22.95, 22.70 mL. Mean titre?

22.68 mL (22.65 and 22.70)

4. Expected titre: 25.0 mL of 0.120 M HCl with 0.100 M NaOH?

30.0 mL

5. pH of 0.050 M HCl?

1.30

6. pH at half-equivalence for formic acid?

about 3.75

7. 0.2042 g KHP needs 20.00 mL NaOH. c(NaOH)?

0.05000 mol/L

8. 0.250 g monoprotic acid needs 20.5 mL of 0.100 M NaOH. Molar mass?

122 g/mol

9. Double titration: V1 12.0, V2 12.0 mL of 0.1 M HCl, 25 mL sample. Composition?

0.048 M Na2CO3 only

10. 12.0 mL of 0.1 N NaOH for 10 mL milk. Lactic acid %?

1.08%

14.5 Mistakes that lose marks

  • Averaging the rough titre.
  • Forgetting to convert mL to L.
  • Ignoring the mole ratio.
  • Recording burette readings to 0.1 mL instead of 0.05 mL.
  • Too many significant figures.
  • Wrong indicator for a weak acid or base.

14.6 Check-yourself and lab questions

Check: Why exclude the rough titre?

It is run fast to locate the endpoint and usually overshoots.

Check: Why is pH at equivalence above 7 for acetic acid?

The acetate ion is a weak base.

Check: What is the pH at half-equivalence for a weak acid?

pH = pKa.

Check: Why does H2SO4 need twice as much NaOH?

It has two acidic protons.

Lab: Why rinse the burette with titrant?

Water left inside would dilute the titrant.

Lab: Why not rinse the conical flask with analyte?

It would add extra analyte; water is fine because moles do not change.

Lab: Why read the bottom of the meniscus at eye level?

To avoid parallax error.

Lab: Why swirl constantly near the endpoint?

So the titrant mixes before the next drop.

✅ 15. Which Titration Method Should You Use?

SituationMethodCalculator modeExample
Unknown acid or base concentrationDirect titrationConcentration0.09820 M HCl
Titrant concentration not certainStandardize against KHPConcentration (titrant)0.1003 M NaOH
Choosing an indicator or reading pKapH curvepH titration curveacetic acid, pH 8.73
Identifying an unknown acidMolar massMolar mass118.1 g/mol
Insoluble or slow sampleBack titrationBack titrationCaCO₃ in antacid
Mixture of alkalisTwo indicatorsDouble titrationNaOH + Na₂CO₃
Food and beverage acidTitratable acidityTitratable acidity1.00% citric
Indicator titration works when
  • The pH jump at equivalence is steep (2 or more pH units).
  • An indicator’s range sits inside the jump.
  • The solution is not strongly colored.
Use a pH meter when
  • The acid or base is weak and dilute.
  • Two equivalence points are close together.
  • The sample is colored or turbid (wine, juice; Boulton, 1980).

🔧 16. Common Titration Calculation Mistakes

Almost every wrong titration answer comes from volumes, ratios or the endpoint. The titration calculator guards against each one below.

SymptomCauseFix
Answer 1000 times too bigUsed mL instead of LDivide mL by 1000
H₂SO₄ concentration doubledIgnored the 1 : 2 ration(acid) = n(NaOH) ÷ 2
Mean titre too highRough titre includedExclude it; average concordant titres
Titres scattered by 0.5 mLOvershooting, air bubble, parallaxAdd dropwise near the end; read at eye level
Pink color fadesCO₂ absorbed by NaOHTake the first permanent pink (about 30 s)
Wrong indicatorRange outside the pH jumpUse the pH curve mode to choose
Molar mass half the expectedDiprotic acid treated as monoproticSet the number of acidic protons
Back titration gives negative analyteNot enough excess reagentAdd more reagent; repeat
Double titration nonsenseV2 entered as total, not extraV2 is the volume after the first endpoint
Titratable acidity differs by acidDifferent meq weightsReport which acid it is expressed as

⚠️ 17. Assumptions and Limitations

AssumptionHow to checkIf it fails
Reaction is complete and fastSharp, stable endpointBack titrate, or heat
Mole ratio from the equationBalanced equationRebalance; check the product formed at the endpoint
Endpoint equals equivalence pointIndicator range in the jumpUse a pH meter and derivative
Concentrations are not activitiesBelow about 0.1 mol/LpH curves shift slightly at high ionic strength
25 °C and Kw = 1.0 × 10⁻¹⁴Room temperatureEquivalence pH changes with temperature
Typical pKa valuesLiterature (Goldberg et al., 2002)Enter your own pKa
Tolerances are rectangularClass A glassware specificationsUse measured repeatability

Ionic equilibria at higher concentrations, activity corrections and mixed solvents are covered in specialist texts (Burgot, 2012).

🎯 18. Conclusion

Your result

The titration calculator gives 0.09820 mol/L from a mean titre of 24.55 mL.

What the titration calculator does

Every titration rests on moles = concentration × volume and a mole ratio. This titration calculator applies that with concordant titres and uncertainty, standardizes titrants, predicts endpoints, draws exact pH curves with equivalence points and indicators, and handles molar mass, back titration, double titration and titratable acidity, with every step shown.

Watch out for

  • Exclude the rough titre.
  • Convert mL to L.
  • Use the mole ratio.
  • Match the indicator to the pH jump.
  • State which acid titratable acidity is expressed as.

Next steps

To make up titrant and standard solutions, use the molarity calculator and the dilution calculator; for EDTA titrations of calcium and magnesium, the water hardness calculator. Whatever the reaction, start and finish with the titration calculator.

❓ 19. Frequently Asked Questions

Q1. How do you calculate concentration from titration?

Multiply the titrant concentration by the mean titre in liters to get moles of titrant. Use the mole ratio from the balanced equation to convert to moles of analyte, then divide by the analyte volume in liters. With 24.55 mL of 0.1000 M NaOH and a 25.00 mL HCl aliquot, c(HCl) = 0.09820 mol/L.

Q2. How do you calculate molarity from titration data?

Molarity of the unknown = (molarity of titrant × titre × a/b) ÷ volume of unknown, where a/b is the analyte-to-titrant mole ratio. Use the same volume units top and bottom, or convert both to liters. For a 1 : 1 reaction the formula reduces to c1V1 = c2V2, which is why many students remember it that way.

Q3. How do you calculate the average titre?

Discard the rough (first) titration, which only finds the approximate endpoint. Then average the titres that agree within 0.10 mL of each other. From 24.80, 24.55, 24.60 and 24.50 mL, the concordant titres are 24.50, 24.55 and 24.60, so the mean titre is 24.55 mL. A titre that does not agree is left out.

Q4. What does concordant titres mean?

Concordant titres are results that agree closely, usually within 0.10 mL, which is two drops from a burette. They show the endpoint was reproducible. Exam boards expect you to average only concordant titres. If none agree, repeat the titration rather than averaging scattered values, because the average would hide an error.

Q5. How do you calculate moles in a titration?

Moles = concentration in mol/L × volume in liters. For 24.55 mL of 0.1000 M NaOH, moles = 0.1000 × 0.02455 = 0.002455 mol. Then use the balanced equation to find moles of the other reactant: the same for HCl, half as many for H2SO4, and five times as many for Fe2+ reacting with permanganate.

Q6. How do you calculate the equivalence point of a titration?

The equivalence point volume equals the moles of analyte times the mole ratio, divided by the titrant concentration. For 25.0 mL of 0.0800 M HCl titrated with 0.100 M NaOH, moles of HCl = 0.00200, so the equivalence point is at 20.0 mL. The pH at that point depends on the acid and base involved.

Q7. How do you calculate the pH at the equivalence point?

At equivalence only the salt remains. For a strong acid with a strong base, pH = 7. For a weak acid with a strong base, the conjugate base hydrolyzes and pH is above 7: 0.1 M acetic acid with 0.1 M NaOH gives pH 8.73. For a weak base with a strong acid, pH is below 7: ammonia gives 5.28.

Q8. How do you calculate the initial pH of a titration?

For a strong acid, pH = −log c: 0.1 M HCl starts at pH 1.00. For a weak acid, solve Ka = x² ÷ (c − x), or use pH ≈ ½(pKa − log c) when the acid is weak and not too dilute: 0.1 M acetic acid starts at pH 2.88. The calculator solves the exact equation for any case.

Q9. How do you calculate pH during a titration?

Before equivalence for a weak acid, the solution is a buffer and pH = pKa + log([A⁻]/[HA]). At half-equivalence pH equals pKa. After equivalence, pH comes from the excess titrant. This calculator solves the full charge balance at every volume, so the same method is exact in every region, including polyprotic acids.

Q10. How do you calculate pH after the equivalence point?

After equivalence, excess strong titrant controls the pH. Find the excess moles, divide by the total volume, and convert. For 25 mL of 0.1 M HCl after adding 30 mL of 0.1 M NaOH, excess OH⁻ = 0.0005 mol in 55 mL = 0.00909 M, so pOH = 2.04 and pH = 11.96.

Q11. Which indicator should you use for a titration?

Choose an indicator whose color-change range lies on the steep part of the curve. Strong acid with strong base: phenolphthalein or methyl orange both work. Weak acid with strong base: phenolphthalein (8.2 to 10). Weak base with strong acid: methyl red (4.4 to 6.2) or methyl orange. Weak with weak: use a pH meter.

Q12. How do you calculate molar mass from titration?

Find the moles of base used, divide by the number of acidic protons to get moles of acid, then divide the mass of acid by those moles. For an aliquot, scale the moles by flask volume ÷ aliquot volume first. 0.1500 g of a diprotic acid needing 25.40 mL of 0.1000 M NaOH gives 118.1 g/mol.

Q13. How do you do titration calculations for sulfuric acid?

Write the equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O. Moles of H2SO4 are half the moles of NaOH. With a mean titre of 22.13 mL of 0.100 M NaOH for 25.0 mL of acid, NaOH = 0.002213 mol, H2SO4 = 0.001106 mol, and the concentration is 0.0443 mol/L. Forgetting the ratio doubles the answer.

Q14. How do you calculate a back titration?

Add a known excess of reagent, let it react with the sample, then titrate the leftover reagent. Moles that reacted = moles added − moles left over. Convert with the mole ratio. With 50.00 mmol HCl added and 20.00 mmol left over, 30.00 mmol reacted, which is 15.00 mmol or 1.501 g of CaCO3.

Q15. How do you calculate a double titration?

Titrate with HCl to the phenolphthalein endpoint (V1), then add methyl orange and continue (V2 extra). If V1 is larger than V2, the mixture is NaOH and Na2CO3: carbonate = c × V2 and hydroxide = c × (V1 − V2). If V2 is larger, it is Na2CO3 and NaHCO3: carbonate = c × V1, bicarbonate = c × (V2 − V1).

Q16. How do you calculate titratable acidity?

Titratable acidity (%) = mL of NaOH × normality × milliequivalent weight of the acid ÷ sample mass or volume × 100. Use 0.06404 g/mEq for citric acid, 0.06704 for malic, 0.07504 for tartaric, 0.09008 for lactic and 0.06005 for acetic acid. 15.60 mL of 0.1000 N NaOH for 10.0 mL of juice is 1.00% citric acid.

Q17. How do you standardize NaOH with KHP?

Weigh potassium hydrogen phthalate, a primary standard with a molar mass of 204.22 g/mol, dissolve it and titrate with the NaOH to phenolphthalein. Moles of KHP = mass ÷ 204.22, which equal moles of NaOH. Divide by the titre in liters. 0.5105 g KHP needing 25.00 mL of NaOH gives 0.1000 mol/L.

Q18. How do you calculate the expected endpoint before a titration?

Estimate the analyte concentration, then calculate expected titre = analyte concentration × analyte volume × b/a ÷ titrant concentration. Knowing it lets you run quickly to about 1 mL before and then add dropwise. If the expected titre is over 50 mL, use a smaller aliquot or a more concentrated titrant so it fits a single burette fill.

Q19. What is the uncertainty of a titration result?

Combine the relative uncertainties of the titre, the pipette volume and the titrant concentration, plus the repeatability of the concordant titres. With a ±0.05 mL burette, ±0.03 mL pipette and ±0.0002 M titrant, the relative standard uncertainty for the headline example is about 0.21%, giving 0.09820 ± 0.00042 mol/L at about 95% confidence.

Q20. How do you do titration calculations in Excel?

Put titrant molarity in A2, the aliquot volume in B2 and the concordant titres in C2 to E2. The concentration is =A2*AVERAGE(C2:E2)/B2 for a 1 : 1 reaction; multiply by a/b for other ratios. For moles, use =A2*AVERAGE(C2:E2)/1000. Use =MAX(C2:E2)-MIN(C2:E2) to check that the titres are concordant, and fill the formulas down for more samples.

📑 20. Cite This Tool

APA 7th edition
StatsUnlock. (2026). Titration calculator [Interactive laboratory tool]. https://statsunlock.com/titration-calculator/
BibTeX
@misc{statsunlock_titration_calculator_2026, title={Titration Calculator}, author={{StatsUnlock Team}}, year={2026}, note={Interactive laboratory tool}, url={https://statsunlock.com/titration-calculator/}}
Methods sentence (fills from your last calculation)
Concentrations were calculated from the mean of concordant titres and the reaction stoichiometry, with GUM uncertainty propagation, using the StatsUnlock Titration Calculator (2026).

🔗 21. Related Tools

📖 22. Glossary of Terms

TermPlain-English meaning
AliquotA measured portion of a solution, usually by pipette.
AnalyteThe substance whose amount is being found.
Back titrationTitrating the excess of a reagent added to the sample.
BuretteGraduated tube that delivers the titrant, read to 0.05 mL.
Concordant titresTitres that agree within about 0.10 mL.
Double titrationTwo endpoints with two indicators to analyze a mixture.
End pointWhere the indicator changes color.
Equivalence pointWhere reactants have reacted in exact stoichiometric amounts.
Gran plotLinear plot used to find the equivalence point from pH data.
Half-equivalence pointHalf-way to equivalence; pH = pKa for a weak acid.
IndicatorWeak acid or base that changes color over a pH range.
KHPPotassium hydrogen phthalate, a primary standard for bases.
Mole ratioRatio of moles from the balanced equation.
NormalityEquivalents per liter; molarity × equivalents per mole.
pKa−log of the acid dissociation constant.
Primary standardPure, stable substance used to standardize a solution.
Rough titreFirst fast titration to find the approximate endpoint.
StandardizationFinding a titrant’s exact concentration.
TitrantSolution of known concentration in the burette.
Titratable acidityTotal acid neutralized by base, expressed as one acid.

📚 23. References

Peer-reviewed sources behind this titration calculator: exact acid–base titration curves, Gran plots and equivalence-point uncertainty, pKa data, carbonate alkalinity, the Kjeldahl titration, EDTA titrations, titratable acidity of foods and wine, antacid back titration, student understanding of titrations, atomic weights and measurement uncertainty. Standards (JCGM 100:2008, ISO 385 and ISO 648 glassware classes) are named where they are used. Every DOI checked October 2026.

  1. Asuero, A. G., & Michałowski, T. (2011). Comprehensive formulation of titration curves for complex acid-base systems and its analytical implications. Critical Reviews in Analytical Chemistry, 41(2), 151–187. https://doi.org/10.1080/10408347.2011.559440
  2. Ayensu, I., Bekoe, S. O., Adu, J. K., Brobbey, A. A., & Appiah, E. (2020). Evaluation of acid neutralizing and buffering capacities of selected antacids in Ghana. Scientific African, 8, e00347. https://doi.org/10.1016/j.sciaf.2020.e00347
  3. Boulton, R. (1980). The relationships between total acidity, titratable acidity and pH in wine. American Journal of Enology and Viticulture, 31(1), 76–80. https://doi.org/10.5344/ajev.1980.31.1.76
  4. Burgot, J.-L. (2012). Ionic equilibria in analytical chemistry. Springer. https://doi.org/10.1007/978-1-4419-8382-4
  5. Cole, T. J. (2015). Too many digits: The presentation of numerical data. Archives of Disease in Childhood, 100(7), 608–609. https://doi.org/10.1136/archdischild-2014-307149
  6. de Levie, R. (1993). Explicit expressions of the general form of the titration curve in terms of concentration: Writing a single closed-form expression for the titration curve for a variety of titrations without using approximations or segmentation. Journal of Chemical Education, 70(3), 209. https://doi.org/10.1021/ed070p209
  7. de Levie, R. (1999). A general simulator for acid-base titrations. Journal of Chemical Education, 76(7), 987. https://doi.org/10.1021/ed076p987
  8. Goldberg, D. (1991). What every computer scientist should know about floating-point arithmetic. ACM Computing Surveys, 23(1), 5–48. https://doi.org/10.1145/103162.103163
  9. Goldberg, R. N., Kishore, N., & Lennen, R. M. (2002). Thermodynamic quantities for the ionization reactions of buffers. Journal of Physical and Chemical Reference Data, 31(2), 231–370. https://doi.org/10.1063/1.1416902
  10. Gran, G. (1952). Determination of the equivalence point in potentiometric titrations. Part II. The Analyst, 77(920), 661–671. https://doi.org/10.1039/an9527700661
  11. Kragten, J. (1994). Calculating standard deviations and confidence intervals with a universally applicable spreadsheet technique. Analyst, 119(10), 2161–2165. https://doi.org/10.1039/AN9941902161
  12. Michałowski, T., & Asuero, A. G. (2012). New approaches in modeling carbonate alkalinity and total alkalinity. Critical Reviews in Analytical Chemistry, 42(3), 220–244. https://doi.org/10.1080/10408347.2012.660067
  13. Michałowski, T., Asuero, A. G., & Wybraniec, S. (2013). The titration in the Kjeldahl method of nitrogen determination: Base or acid as titrant? Journal of Chemical Education, 90(2), 191–197. https://doi.org/10.1021/ed200863p
  14. Prohaska, T., Irrgeher, J., Benefield, J., Böhlke, J. K., Chesson, L. A., Coplen, T. B., Ding, T., Dunn, P. J. H., Gröning, M., Holden, N. E., Meijer, H. A. J., Moossen, H., Possolo, A., Takahashi, Y., Vogl, J., Walczyk, T., Wang, J., Wieser, M. E., Yoneda, S., Zhu, X.-K., & Meija, J. (2022). Standard atomic weights of the elements 2021 (IUPAC Technical Report). Pure and Applied Chemistry, 94(5), 573–600. https://doi.org/10.1515/pac-2019-0603
  15. Rossotti, F. J. C., & Rossotti, H. (1965). Potentiometric titrations using Gran plots: A textbook omission. Journal of Chemical Education, 42(7), 375. https://doi.org/10.1021/ed042p375
  16. Sadler, G. D., & Murphy, P. A. (2010). pH and titratable acidity. In S. S. Nielsen (Ed.), Food analysis (4th ed., pp. 219–238). Springer. https://doi.org/10.1007/978-1-4419-1478-1_13
  17. Schwartz, L. M. (1992). Uncertainty of a titration equivalence point: A graphical method using spreadsheets to predict values and detect systematic errors. Journal of Chemical Education, 69(11), 879. https://doi.org/10.1021/ed069p879
  18. Sheppard, K. (2006). High school students’ understanding of titrations and related acid-base phenomena. Chemistry Education Research and Practice, 7(1), 32–45. https://doi.org/10.1039/B5RP90014J
  19. Tyl, C., & Sadler, G. D. (2017). pH and titratable acidity. In S. S. Nielsen (Ed.), Food analysis (5th ed., pp. 389–406). Springer. https://doi.org/10.1007/978-3-319-45776-5_22
  20. Yappert, M. C., & DuPré, D. B. (1997). Complexometric titrations: Competition of complexing agents in the determination of water hardness with EDTA. Journal of Chemical Education, 74(12), 1422. https://doi.org/10.1021/ed074p1422
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