A 1.0 -L saturated silver carbonate solution at \(5^{\circ} \mathrm{C}\) is treated with enough hydrochloric acid to decompose the compound. The carbon dioxide generated is collected in a \(19-\mathrm{mL}\) vial and exerts a pressure of \(114 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C}\). What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(5^{\circ} \mathrm{C} ?\)

When hydrochloric acid is added to the saturated solution of silver carbonate, it decomposes to yield silver chloride and carbon dioxide. The balanced decomposition reaction is: \[2 HCl + Ag_{2}CO_{3} \rightarrow 2 AgCl + CO_{2} + H_{2}O\] After the decomposition, the dissolution of \(Ag_{2}CO_{3}\) in water can be represented by this equilibrium equation: \[Ag_{2}CO_{3} \leftrightarrows 2 Ag^{+} + CO_{3}^{2-}\] These equations will give insight into the stoichiometric relationship between the reactants and products.

As per the ideal gas law, \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. From the given information, the pressure \(P\) is 114 mmHg, which should be converted to atm by dividing by 760: \(P = 114/760 = 0.15\) atm. The volume \(V\) is 19 mL, which must be converted to L by dividing by 1000: \(V = 19/1000 = 0.019\) L. The temperature \(T\) is \(25^{\circ} C\), which must be converted to K by adding 273.15: \(T = 25 + 273.15 = 298.15 K\). The gas constant \(R = 0.0821\) atm.L/K.mol. Rearranging the gas law equation to solve for n yields the number of moles of CO2: \(n = PV/RT = (0.15 atm * 0.019 L) / (0.0821 atm.L/K.mol * 298.15 K) = 7.71 * 10^{-4} mol\)

Looking back at the balanced reaction, 1 mol of \(Ag_{2}CO_{3}\) yields 1 mol of CO2, so the molar amount of \(Ag_{2}CO_{3}\) that decomposed must be the same as the moles of CO2 formed: \(7.71 * 10^{-4}\) mol.

As 1 mol of \(Ag_{2}CO_{3}\) yields 2 mol of \(Ag^{+}\) and 1 mol of \(CO_{3}^{2-}\), the molar concentration of \(Ag^{+}\) ions is \(2 * 7.71 * 10^{-4}\) mol/L = \(1.54 * 10^{-3}\) M, and of \(CO_{3}^{2-}\) ions is \(7.71 * 10^{-4}\) M. The \(K_{sp}\) is then the product of the molar concentrations of these ions at equilibrium: \(K_{sp} = [Ag^{+}]^2 * [CO_{3}^{2-}] = (1.54 * 10^{-3})^2 * 7.71 * 10^{-4} = 1.82 * 10^{-9}\)