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Difficulty: MediumAcid-Base Titrations, Indicators, and Volumetric Calculations

In an acid-base titration, 25.0 cm325.0\text{ cm}^3 of a 0.10 mol dm−30.10\text{ mol dm}^{-3} sodium hydroxide (NaOH\text{NaOH}) solution required 12.5 cm312.5\text{ cm}^3 of a tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) solution for complete neutralization. What is the concentration of the acid solution in g dm−3\text{g dm}^{-3}? [H=1.0,O=16.0,S=32.0][\text{H} = 1.0, \text{O} = 16.0, \text{S} = 32.0]

  1. 9.80 g dm−39.80\text{ g dm}^{-3}Answer
  2. B
    19.60 g dm−319.60\text{ g dm}^{-3}
  3. C
    4.90 g dm−34.90\text{ g dm}^{-3}
  4. D
    0.10 g dm−30.10\text{ g dm}^{-3}

Answer

The concentration of the tetraoxosulfate(VI) acid solution is 9.80 g dm−39.80\text{ g dm}^{-3}.
The balanced chemical equation H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} establishes a mole ratio (na:nbn_a : n_b) of 1:21 : 2. Substituting the given values into CaVaCbVb=nanb\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} gives Ca=0.10 mol dm−3C_a = 0.10\text{ mol dm}^{-3}. Multiplying this by the molar mass of H2SO4\text{H}_2\text{SO}_4 (98.0 g mol−198.0\text{ g mol}^{-1}) yields 9.80 g dm−39.80\text{ g dm}^{-3}.

Step-by-Step Solution

1
Write the balanced chemical equation to determine the mole ratio.
H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}, giving na=1n_a = 1 and nb=2n_b = 2.
Stoichiometric coefficients define the mole ratio required for complete neutralization.
2
Calculate the molar concentration of tetraoxosulfate(VI) acid (CaC_a) using the titration formula CaVaCbVb=nanb\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}.
Ca×12.50.10×25.0=12  ⟹  Ca×12.5=1.25  ⟹  Ca=0.10 mol dm−3\frac{C_a \times 12.5}{0.10 \times 25.0} = \frac{1}{2} \implies C_a \times 12.5 = 1.25 \implies C_a = 0.10\text{ mol dm}^{-3}.
Relates the volumes and concentrations of acid and base according to their stoichiometry.
3
Convert the molar concentration to mass concentration in g dm−3\text{g dm}^{-3}.
Molar mass of H2SO4=2(1.0)+32.0+4(16.0)=98.0 g mol−1\text{H}_2\text{SO}_4 = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-1}. Mass concentration =0.10 mol dm−3×98.0 g mol−1=9.80 g dm−3= 0.10\text{ mol dm}^{-3} \times 98.0\text{ g mol}^{-1} = 9.80\text{ g dm}^{-3}.
Mass concentration equals molar concentration multiplied by relative molar mass.

Key Concept

Volumetric calculations involving diprotic acid-monoprotic base titrations and conversion between molarity and mass concentration.
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Topics

Acid-Base Titrations, Indicators, and Volumetric Calculations

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