CAIE A-Level Chemistry 2.4.1 Reacting-Mass and Volume Calculations

Practise a consistent mole-ratio method across reacting masses, gases, solutions, yields and reagent excess.

[Maximum number: 3]

Bismuth is an element in Group 15 of the Periodic Table.

$\mathrm{NaBiO}_{3}$ can be used to determine the concentration of $\mathrm{Mn}^{2+}(\mathrm{aq})$ . The ionic equation for the reaction is shown in equation 2 .

$equation $2 \quad 2 \mathrm{Mn}^{2+}+5 \mathrm{BiO}_{3}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$$

A student uses the following procedure in an experiment.
- Add $100.0 \mathrm{~cm}^{3}$ of a saturated solution of $\mathrm{Mn}^{2+}(\mathrm{aq})$ to a volumetric flask.
- Add distilled water to the flask to make a $1.00 \mathrm{dm}^{3}$ diluted solution.
- Titrate a $25.00 \mathrm{~cm}^{3}$ sample of the diluted solution with $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaBiO}_{3}(\mathrm{aq})$ .

The $25.00 \mathrm{~cm}^{3}$ sample of the diluted solution of $\mathrm{Mn}^{2+}(\mathrm{aq})$ reacts completely with exactly $21.50 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaBiO}_{3}(\mathrm{aq})$ .

Calculate the concentration, in $\mathrm{moldm}^{-3}$ , of $\mathrm{Mn}^{2+}(\mathrm{aq})$ in the saturated solution.
Show your working.

concentration of $\mathrm{Mn}^{2+}(\mathrm{aq})$ in the saturated solution = $\mathrm{mol} \mathrm{dm}^{-3}$

moles of $\mathrm{NaBiO}_{3}=21.50 / 1000 \times 0.100 \quad\left(=2.15 \times 10^{-3} \mathrm{~mol}\right)$

moles of $\mathrm{Mn}^{2+}$ in dil. $=2.15 \times 10^{-3} \times 2 / 5 \times 40(=0.0344 \mathrm{~mol})$

conc. saturated $\mathrm{Mn}^{2+}=0.0344 \times 1000 / 100=\mathbf{0 . 3 4 4 ~ m o l ~ d m}{ }^{-3}$

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2.4.1—Perform calculations question 1