Question

过量agcl溶解在下列各种物质中，问哪种溶液中c(ag^{+})最小
a 100ml 0.2mol·l^{-1}kcl溶液
b 1000ml 0.5mol·l^{-1}kno_{3}溶液
c 100ml水
d 1000ml水

Explanation:

Step1: Write the solubility product constant expression

For \(AgCl\), \(K_{sp} = c(Ag^{+})\times c(Cl^{-})\). In a solution, the presence of \(Cl^{-}\) (common - ion effect) will suppress the dissolution of \(AgCl\).

Step2: Analyze each option

• Option A: \(100mL\ 0.2mol\cdot L^{-1}KCl\) solution. \(c(Cl^{-})=0.2mol/L\). According to \(K_{sp} = c(Ag^{+})\times c(Cl^{-})\), \(c(Ag^{+})=\frac{K_{sp}}{c(Cl^{-})}\).
• Option B: \(1000mL\ 0.5mol\cdot L^{-1}KNO_{3}\) solution. \(KNO_{3}\) has no common - ion with \(AgCl\). The solubility of \(AgCl\) is relatively larger, and \(c(Ag^{+})\) is larger than in the solution with common - ion.
• Option C: \(100mL\) water. Without any common - ion, the solubility of \(AgCl\) is larger than in the solution with \(Cl^{-}\) (common - ion).
• Option D: \(1000mL\) water. Similar to option C, without common - ion, \(c(Ag^{+})\) is larger than in the solution with \(Cl^{-}\) (common - ion).

Since \(K_{sp}\) of \(AgCl\) is a constant (\(K_{sp}(AgCl)=1.8\times 10^{-10}\) at \(25^{\circ}C\)), and \(c(Cl^{-})\) in option A is larger among the given options (except for no common - ion cases), according to \(c(Ag^{+})=\frac{K_{sp}}{c(Cl^{-})}\), the larger \(c(Cl^{-})\), the smaller \(c(Ag^{+})\).

Answer:

A. \(100mL\ 0.2mol\cdot L^{-1}KCl\) solution