Titration Calculator: Solve for Molarity, Volume, or Moles

Solve an acid-base titration for an unknown molarity, volume, or moles at the equivalence point, with polyprotic acids and bases handled through an explicit acid-to-base mole ratio.

Solve for

Solved value

Enter three values

Leave one of M_A, V_A, M_B, V_B blank and the tool solves it from M_A·V_A / a = M_B·V_B / b.

Titration summary
QuantityValue
Export

M_A · V_A / a = M_B · V_B / b How?

How this is calculated

At the equivalence point the acid and base have reacted in the ratio their balanced equation sets. The coefficients a and b are those of the acid and base in that equation, and each divides its own side of M·V. For H₂SO₄ + 2 NaOH that is a = 1 and b = 2, so the base supplies twice the moles. The tool converts your inputs to mol/L and litres, solves the one blank field, then converts the answer back to your unit.

Formula: M_A · V_A / a = M_B · V_B / b

How to calculate a titration

A titration measures an unknown concentration by reacting it to the equivalence point with a standard solution of known concentration. Once you know the acid concentration and volume and the base concentration and volume, three of those four numbers fix the fourth. Enter the three you measured, leave the unknown blank, and the tool returns it along with the moles of analyte and the reaction ratio it used.

Take a monoprotic case at a 1:1 ratio. Titrating 25.00 mL of an acid to the endpoint takes 32.00 mL of 0.100 M base. With a = 1 and b = 1 the balance is M_A · V_A = M_B · V_B, so M_A = 0.100 · 32.00 / 25.00 = 0.128 M. The volume unit cancels on both sides, which is why a titre read in millilitres returns an answer without any unit juggling.

Titration calculations for polyprotic acids and bases

Polyprotic systems are where the numerator (normality) shortcut quietly goes wrong. Sulfuric acid neutralizing sodium hydroxide runs H₂SO₄ + 2 NaOH = Na₂SO₄ + 2 H₂O, so a = 1 and b = 2 and the equivalence condition is M_A · V_A / 1 = M_B · V_B / 2. Solving for the base volume with M_A = 0.100 M, V_A = 25.00 mL and M_B = 0.200 M gives V_B = 2 · 0.100 · 25.00 / 0.200 = 25.00 mL. Reading the equation directly, rather than placing a valence factor in the numerator, keeps the diprotic and triprotic endpoints correct. Pick the ratio that names your endpoint; phosphoric acid has three, and the tool never assumes full neutralization for you.

Reading the moles and the residual

Alongside the solved value the tool reports the moles on each side, n_A = M_A · V_A and n_B = M_B · V_B in litres, so a titration-calculations question that asks for amount of substance is answered in the same pass. When you supply all four values instead of leaving one blank, the tool switches to a check: it reports how far the four numbers sit from the equivalence-point ratio as a percent residual, and flags anything past 1 percent.

Sources

  1. Introductory Chemistry (CK-12) 21.18: Titration Calculations. LibreTexts. Retrieved .
  2. CK-12 Chemistry FlexBook 2.0 21.18: Titration Calculations. CK-12. Retrieved .