CAIE IGCSE Chemistry: Titration Calculations Question Bank

Practise using titration volumes, concentrations and equations to calculate moles of solute and solution concentrations.

[Maximum number: 5]

Solid sodium hydroxide is a base which dissolves to form an aqueous solution, NaOH(aq).

In a titration, $25.0 \mathrm{~cm}^{3}$ of $0.0800 \mathrm{~mol} / \mathrm{dm}^{3}$ aqueous potassium hydroxide, KOH(aq), is neutralised by $20.0 \mathrm{~cm}^{3}$ of dilute sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})$ .

2 \mathrm{KOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})

Calculate the concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ , in $\mathrm{g} / \mathrm{dm}^{3}$ using the following steps.
- Calculate the number of moles of KOH used. mol
- Determine the number of moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ which react with the KOH . mol
- Calculate the concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $\mathrm{mol} / \mathrm{dm}^{3}$ . $\mathrm{mol} / \mathrm{dm}^{3}$
- Calculate the concentration of $\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{in} \mathrm{g} / \mathrm{dm}^{3}$ . $\mathrm{g} / \mathrm{dm}^{3}$
[5]
[Total: 21]

$\begin{aligned} \mathrm{M} 1 \mathrm{~mol} \mathrm{KOH}=0.0800 \times 25 / 1000 =0.002(00) / 2(00) \times 10^{-3} \mathrm{M} 2 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}=\mathrm{M} 1 / 2=0.002 / 2 =0.001(00) / 1\left(00 \times 10^{-3}\right. \end{aligned}$ $\mathrm{M} 3=\mathrm{M} 2 \times 1000 / 20=0.001 \times 1000 / 20$ $=0.05(00) / 5.(00) \times 10^{-2}$
M4 =98
$\text { M5 }=98 \times \text { M3 }=98 \times 0.05(00)=4.9(0)\left(\mathrm{g} / \mathrm{dm}^{3}\right)$

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3.3.6—Experimental data from a titration to question 1