Chemistry · General chemistry II · Worked example

Titrate sulfuric acid, a diprotic acid

A 20.00 mL sample of sulfuric acid is titrated with 0.1250 M NaOH, and the equivalence point is reached after 31.84 mL. Find the concentration of the acid.

Balance the equation

Each H₂SO₄ gives up two protons, so it takes two NaOH.

H_{2} {SO}_{4} (aq) + 2 NaOH (aq) \to {Na}_{2} {SO}_{4} (aq) + 2 H_{2} O (l)

\begin{array}{c} H_{2} {SO}_{4} (aq) + 2 NaOH (aq) \\ \to {Na}_{2} {SO}_{4} (aq) + 2 H_{2} O (l) \end{array}

Moles of base

Volume in liters times molarity.

n = (0.03184 L) (0.1250 mol/L) = 3.980 \times 10^{- 3} mol

\begin{array}{c} n = (0.03184 L) (0.1250 mol/L) \\ = 3.980 \times 10^{- 3} mol \end{array}

Moles of acid

The equation gives 1 mol of H₂SO₄ for every 2 mol of NaOH, so the acid is half the base.

3.980 \times 10^{- 3} mol \times \frac{1}{2} = 1.990 \times 10^{- 3} mol

\begin{array}{c} 3.980 \times 10^{- 3} mol \times \frac{1}{2} \\ = 1.990 \times 10^{- 3} mol \end{array}

Concentration

Divide by the acid’s volume in liters.

c = \frac{1.990 \times 10^{- 3} mol}{0.02000 L} = 0.09950 M

Why M₁V₁ = M₂V₂ fails here

Treating the reaction as 1 : 1 would give 0.1990 M, twice the true value. The shortcut hides the mole ratio, so use it only when the ratio is 1 : 1.

Result

The sulfuric acid is 0.09950 M.

Your turn

What volume of 0.200 M HCl neutralizes 25.0 mL of 0.150 M Ba(OH)₂?

Show the answer and explanation

37.5 mL.

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O. The base is (0.0250 L)(0.150 mol/L) = 3.75 × 10⁻³ mol, so the acid is 7.50 × 10⁻³ mol, and 7.50 × 10⁻³ mol ÷ 0.200 mol/L = 0.0375 L.

\frac{7.50 \times 10^{- 3}}{0.200} = 0.0375

Keep exploring

Stoichiometry & yield opens with 31.84 mL of 0.1250 M NaOH and the acid in excess. It shows 3.980 × 10⁻³ mol of base consuming 1.990 × 10⁻³ mol of H₂SO₄.

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