To precipitate as \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s}),\) all the \(\mathrm{Ag}^{+}\) present in \(338 \mathrm{mL}\) of a saturated solution of \(\mathrm{AgBrO}_{3}\) requires \(30.4 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) measured at \(23^{\circ} \mathrm{C}\) and \(748 \mathrm{mm} \mathrm{Hg} .\) What is \(K_{\mathrm{sp}}\) for \(\mathrm{AgBrO}_{3} ?\)

The moles of \(\mathrm{Ag}^{+}\) can be determined by using the given volume of \(\mathrm{H}_{2}\mathrm{S}\) since the reaction between \(\mathrm{Ag}^{+}\) and \(\mathrm{H}_{2}\mathrm{S}\) forms \(\mathrm{Ag}_{2} \mathrm{S}\). Given the Ideal gas law we have the equation \(PV=nRT \Rightarrow n=PV/RT\). Here, \(P\) in Atmospheres will be \(748 \times 1 / 760 = 0.9842 \, atm\), \(V = 30.4 \, mL = 0.03 \, L\), \(R\) is the ideal gas constant = 0.0821 \(L.atm/K.mol\), and \(T\) is the temperature = 23 Celsius = 296.15 Kelvin. Substituting the values, we find number of moles of \(H_2S\) to be \(n = 0.0012 \, mol\). Since it reacts in a 1:2 ratio with \(Ag^{+}\), moles of \(Ag^{+}\) = 2 × moles of \(H_2S\) = 0.0024 mol.

As per the reaction, one molecule of \(\mathrm{AgBrO}_{3}\) gives one ion each of \(Ag^{+}\) and \(BrO_{3}^{-}\). Therefore, concentration of \(Ag^{+}\) will be equal to the concentration of \(BrO_{3}^{-}\). The Volume of \(\mathrm{AgBrO}_{3}\) is 338 mL = 0.338 L. Hence, concentration = moles/volume = 0.0024 / 0.338 = 0.0071 M.

The solubility product constant (\(K_{sp}\)) is given by: \(K_{sp}\) = [\(Ag^{+}\)] × [\(BrO_{3}^{-}\)] = (0.0071) × (0.0071) = 0.00005041.