Calculate the solubility (in grams per liter) of silver chloride in the following. (a) pure water (b) \(0.025 \mathrm{M} \mathrm{BaCl}_{2}\) (c) \(0.17 \mathrm{M} \mathrm{AgNO}_{3}\)

(We need to write the chemical reaction for the dissolution of silver chloride (AgCl). The dissociation of AgCl can be represented as: \(\ce{AgCl (s) <=> Ag+ (aq) + Cl- (aq)}\) The Ksp for \(\ce{AgCl}\) is \(1.6 \times 10^{-10}\) at 25°C.)

(Let 's' denote the solubility of \(\ce{AgCl}\) in water. According to the dissociation reaction, when 's' moles of \(\ce{AgCl}\) dissolve, we get 's' moles of both \(\ce{Ag+}\) and \(\ce{Cl-}\) ions. The Ksp expression for the reaction can be written as: \(K_{sp}=[\ce{Ag^+}][\ce{Cl^-}]\) Since both concentrations are equal to 's', we can write: \(K_{sp} = s^2\) To find the solubility, solve for s: \(s=\sqrt{K_{sp}}=\sqrt{1.6 \times 10^{-10}}\) \(s = 1.26 \times 10^{-5} \, \mathrm{mol/L}\) Convert the solubility to grams per liter: \(1.26 \times 10^{-5} \,\mathrm{mol/L} \times 143.3 \,\mathrm{g/mol} = 1.8 \times 10^{-3} \,\mathrm{g/L}\) So, the solubility of \(\ce{AgCl}\) in pure water is \(1.8 \times 10^{-3} \,\mathrm{g/L}\).)

(In this case, the external source of chloride ions is present due to \(\mathrm{BaCl}_{2}\). So, the concentration of \(\ce{Cl-}\) due to \(\mathrm{BaCl}_{2}\) is \(0.025 \, \mathrm{M}\). The Ksp expression becomes: \(K_{sp} = [\ce{Ag^+}] ([\ce{Cl^-}]_0 + s)\) Solve for s: \(s = \frac{K_{sp}}{[\ce{Cl^-}]_0} = \frac{1.6 \times 10^{-10}}{0.025}\) \(s = 6.4 \times 10^{-9} \, \mathrm{mol/L}\) Convert the solubility to grams per liter: \(6.4 \times 10^{-9} \,\mathrm{mol/L} \times 143.3 \,\mathrm{g/mol} = 9.2 \times 10^{-7} \,\mathrm{g/L}\) So, the solubility of \(\ce{AgCl}\) in \(0.025 \, \mathrm{M} \, \mathrm{BaCl}_{2}\) solution is \(9.2 \times 10^{-7} \,\mathrm{g/L}\).)

(In this case, the external source of silver ions is present due to \(\mathrm{AgNO}_{3}\). So, the concentration of \(\ce{Ag+}\) due to \(\mathrm{AgNO}_{3}\) is \(0.17 \, \mathrm{M}\). The Ksp expression becomes: \(K_{sp} = ([\ce{Ag^+}]_0 + s) [\ce{Cl^-}]\) Solve for s: \(s = \frac{K_{sp}}{[\ce{Ag^+}]_0} = \frac{1.6 \times 10^{-10}}{0.17}\) \(s = 9.41 \times 10^{-10} \, \mathrm{mol/L}\) Convert the solubility to grams per liter: \(9.41 \times 10^{-10} \,\mathrm{mol/L} \times 143.3 \,\mathrm{g/mol} = 1.35 \times 10^{-7} \,\mathrm{g/L}\) So, the solubility of \(\ce{AgCl}\) in \(0.17 \, \mathrm{M} \, \mathrm{AgNO}_{3}\) solution is \(1.35 \times 10^{-7} \,\mathrm{g/L}\).) In summary, the solubility of silver chloride is: (a) \(1.8 \times 10^{-3} \, \mathrm{g/L}\) in pure water (b) \(9.2 \times 10^{-7} \, \mathrm{g/L}\) in \(0.025 \, \mathrm{M} \, \mathrm{BaCl}_{2}\) solution (c) \(1.35 \times 10^{-7} \, \mathrm{g/L}\) in \(0.17 \, \mathrm{M} \, \mathrm{AgNO}_{3}\) solution