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Exercise 8.8 1. 32,5 cm³ of a sulphuric acid solution neutralises exactly 25 cm³ of a potassium hydroxide solution of concentration 0,6 mol.dm⁻³. a) Write a balanced equation for the reaction. b) Name a suitable indicator for the reaction. c) Calculate the concentration of the acid solution.

Chemistry

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Answer

c

Ca=0.231 mol.dm−3C_a = 0.231 \text{ mol.dm}^{-3}

Step-by-step solution

4 steps

a) To write a balanced equation for the reaction between sulphuric acid (H2SO4\text{H}_2\text{SO}_4) and potassium hydroxide (KOH\text{KOH}), we recognize it as an acid-base neutralization reaction, which produces a salt (potassium sulfate) and water.

Step 1: Write the unbalanced equation.

H2SO4(aq)+KOH(aq)→K2SO4(aq)+H2O(l)\text{H}_2\text{SO}_4(\text{aq}) + \text{KOH}(\text{aq}) \rightarrow \text{K}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l})

Step 2: Balance the potassium atoms by placing a coefficient of 2 in front of KOH\text{KOH}.

H2SO4(aq)+2KOH(aq)→K2SO4(aq)+H2O(l)\text{H}_2\text{SO}_4(\text{aq}) + 2\text{KOH}(\text{aq}) \rightarrow \text{K}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l})

Step 3: Balance the hydrogen and oxygen atoms by placing a coefficient of 2 in front of H2O\text{H}_2\text{O}.

H2SO4(aq)+2KOH(aq)→K2SO4(aq)+2H2O(l)\boxed{\text{H}_2\text{SO}_4(\text{aq}) + 2\text{KOH}(\text{aq}) \rightarrow \text{K}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})}

Part b

For a strong acid-strong base titration, the equivalence point is at pH 7. A suitable indicator should have a color change range that includes pH 7.

A suitable indicator for this reaction is phenolphthalein.

Part c

Ca=0.231 mol.dm−3C_a = 0.231 \text{ mol.dm}^{-3}

To calculate the concentration of the acid solution, we use the titration formula based on the mole ratio from the balanced equation.

  1. Identify the given values and the mole ratio.

    Given:
    Volume of acid, Va=32.5 cm3\text{V}_a = 32.5 \text{ cm}^3
    Volume of base, Vb=25 cm3\text{V}_b = 25 \text{ cm}^3
    Concentration of base, Cb=0.6 mol.dm−3\text{C}_b = 0.6 \text{ mol.dm}^{-3}
    From the balanced equation, the mole ratio of acid to base (na:nb\text{n}_a : \text{n}_b) is 1:21:2.

  2. Use the titration formula CaVana=CbVbnb\frac{C_a V_a}{n_a} = \frac{C_b V_b}{n_b} to find the concentration of the acid (Ca\text{C}_a).

    Ca=nanb×CbVbVaC_a = \frac{n_a}{n_b} \times \frac{C_b V_b}{V_a}

  3. Substitute the values into the formula.

    Ca=12×0.6 mol.dm−3×25 cm332.5 cm3C_a = \frac{1}{2} \times \frac{0.6 \text{ mol.dm}^{-3} \times 25 \text{ cm}^3}{32.5 \text{ cm}^3}

  4. Calculate the concentration.

    Ca=12×1532.5 mol.dm−3C_a = \frac{1}{2} \times \frac{15}{32.5} \text{ mol.dm}^{-3} Ca=12×0.461538... mol.dm−3C_a = \frac{1}{2} \times 0.461538... \text{ mol.dm}^{-3} Ca=0.230769... mol.dm−3C_a = 0.230769... \text{ mol.dm}^{-3}

    Rounding to three significant figures:

    Ca=0.231 mol.dm−3\boxed{C_a = 0.231 \text{ mol.dm}^{-3}}
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