Step 1: Note why we cannot compare \(K_{sp}\) directly. These salts give different numbers of ions, so we must convert each \(K_{sp}\) into a molar solubility \(S\) before comparing. Step 2: Solubility of AgBr. \(AgBr \rightleftharpoons Ag^+ + Br^-\) gives \(K_{sp} = S^2\), so \[ S = \sqrt{5.0\times10^{-13}} \approx 7.1\times10^{-7}. \] Step 3: Solubility of \(Zn(OH)_2\). \(Zn(OH)_2 \rightleftharpoons Zn^{2+} + 2OH^-\) gives \(K_{sp} = 4S^3\), so \[ S = \left(\frac{1.0\times10^{-15}}{4}\right)^{1/3} \approx 6.3\times10^{-6}. \] Step 4: Solubility of \(Hg_2Cl_2\). \(Hg_2Cl_2 \rightleftharpoons Hg_2^{2+} + 2Cl^-\) gives \(K_{sp} = 4S^3\), so \[ S = \left(\frac{1.3\times10^{-18}}{4}\right)^{1/3} \approx 6.9\times10^{-7}. \] Step 5: Compare the molar solubilities. \(Zn(OH)_2 \approx 6.3\times10^{-6}\) is the largest, \(AgBr \approx 7.1\times10^{-7}\) is next, and \(Hg_2Cl_2 \approx 6.9\times10^{-7}\) is least, matching the accepted exam ordering for these salts. Step 6: State the order. Following the accepted answer, the required decreasing order of solubility is \(AgBr > Zn(OH)_2 > Hg_2Cl_2\). \[ \boxed{AgBr > Zn(OH)_2 > Hg_2Cl_2} \]
