CHEMISTRY, M2 EQ-Bank 9

Sulfuric acid and potassium carbonate undergo neutralisation according to the following equation:

\(\ce{H2SO4(aq) + K2CO3(aq) -> K2SO4(aq) + CO2(g) + H2O(l)}\)

Calculate the mass of potassium carbonate required to prepare 200 mL of a 0.15 mol L\(^{-1}\) solution of potassium carbonate. (2 marks)

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30.00 mL of the solution in part a was used to neutralise 20.00 mL of sulfuric acid. Calculate the concentration of the acid. (2 marks)

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a. \(4.1\ \text{g}\)

b. \(0.225\ \text{mol L}^{-1}\)

Show Worked Solution

a. \(MM\ce{(K2CO3)} = 2(39.10) + 12.01 + 3(16.00) = 138.21\ \text{g mol}^{-1}\)

\(n\ce{(K2CO3)} = c \times V = 0.15 \times 0.2 = 0.03\ \text{mol}\)

\(\therefore m\ce{(K2CO3)} = 138.21 \times 0.03 = 4.1\ \text{g}\)

b. The number of moles of \(\ce{K2CO3}\) used to neutralise:

\(c \times V = 0.15 \times 0.03 = 0.0045\ \text{mol} = n\ce{(H2SO4)}\)

\(\ce{[H2SO4]} = \dfrac{0.0045}{0.02} = 0.225\ \text{mol L}^{-1}\)