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

During a titration experiment, 25.0 cm325.0\text{ cm}^3 of a potassium hydroxide (KOH\text{KOH}) solution of unknown concentration required 20.0 cm320.0\text{ cm}^3 of a 0.050 mol dm−30.050\text{ mol dm}^{-3} tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) solution for complete neutralization. What is the mass concentration of the potassium hydroxide solution in g dm−3\text{g dm}^{-3}?

[K=39,O=16,H=1\text{K} = 39, \text{O} = 16, \text{H} = 1]

  1. A
    2.24 g dm−32.24\text{ g dm}^{-3}
  2. 4.48 g dm−34.48\text{ g dm}^{-3}Answer
  3. C
    8.96 g dm−38.96\text{ g dm}^{-3}
  4. D
    0.080 g dm−30.080\text{ g dm}^{-3}

Answer

The mass concentration of the potassium hydroxide solution is 4.48 g dm−34.48\text{ g dm}^{-3}.
The reaction between tetraoxosulfate(VI) acid and potassium hydroxide has a 1:21:2 mole ratio (H2SO4+2KOH→K2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}). Substituting the given values into CaVaCbVb=12\frac{C_a V_a}{C_b V_b} = \frac{1}{2} gives Cb=0.080 mol dm−3C_b = 0.080\text{ mol dm}^{-3}. Multiplying this molarity by the molar mass of KOH\text{KOH} (56 g mol−156\text{ g mol}^{-1}) gives the mass concentration of 4.48 g dm−34.48\text{ g dm}^{-3}.

Step-by-Step Solution

1
Write the balanced chemical equation for the neutralization reaction.
H2SO4+2KOH→K2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}
The stoichiometry shows that 1 mole1\text{ mole} of H2SO4\text{H}_2\text{SO}_4 reacts with 2 moles2\text{ moles} of KOH\text{KOH} (na=1,nb=2n_a = 1, n_b = 2).
2
Calculate the molarity (CbC_b) of the potassium hydroxide solution using the titration equation.
CaVaCbVb=nanb  ⟹  0.050×20.0Cb×25.0=12  ⟹  Cb=0.080 mol dm−3\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{0.050 \times 20.0}{C_b \times 25.0} = \frac{1}{2} \implies C_b = 0.080\text{ mol dm}^{-3}
Equating the mole ratio allows determination of the concentration of the base in moles per cubic decimetre.
3
Calculate the molar mass of KOH\text{KOH} and convert the concentration to g dm−3\text{g dm}^{-3}.
Molar mass of KOH=39+16+1=56 g mol−1\text{KOH} = 39 + 16 + 1 = 56\text{ g mol}^{-1}. Mass concentration =0.080 mol dm−3×56 g mol−1=4.48 g dm−3= 0.080\text{ mol dm}^{-3} \times 56\text{ g mol}^{-1} = 4.48\text{ g dm}^{-3}.
Mass concentration is obtained by multiplying molar concentration by the relative molar mass.

Key Concept

Determination of mass concentration from volumetric analysis data using stoichiometric mole ratios.
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Topics

Acid-Base Titrations, Indicators, and Volumetric Calculations

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