If the K_sp for AgCl is equal to 4.65×10^-9 and the K_f for $\mathrm { AgCl } _ { 2 } ^ { - }$ is equal to 8.7×10⁵ at $35 ^ { \circ } \mathrm { C }$ , what is the solubility of AgCl in pure water at the same temperature?

Question

Quizplus · Accepted Answer

We can use the equation for the solubility product constant to find the solubility of AgCl:
$$\mathrm {K_{sp}} = [\mathrm{Ag^{+}}][\mathrm{Cl^{-}}] = s^2$$
$$4.65×10^{-9} = s^2$$
$$s = \sqrt{4.65×10^{-9}} = 6.82×10^{-5}\mathrm{M}$$
Next, we can use the equation for the formation constant of AgCl2- to find the concentration of Ag+ in solution:
$$\mathrm{K_f} = \dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Ag^{+}}][\mathrm{Cl^{-}}]} = \dfrac{8.7\times 10^{15}}{4.65\times 10^{-9}}$$
$$\mathrm{K_f} = 1.8709677\times 10^{24} = [\mathrm{Ag^{+}}]^2$$
$$[\mathrm{Ag^{+}}] = \sqrt{1.8709677\times 10^{24}} = 4.3255\times 10^{12}\mathrm{M}$$
Since AgCl is a strong electrolyte, we can assume that the concentration of Cl- is equal to the solubility of AgCl: 6.82×10^-5M.
Therefore, we can use the solubility product constant equation to find the concentration of Ag+:
$$\mathrm {K_{sp}} = [\mathrm{Ag^{+}}][\mathrm{Cl^{-}}] = (6.82×10^{-5})(4.3255\times 10^{12})$$
$$\mathrm {K_{sp}} = 2.951\times 10^{-6}$$
Thus, the solubility of AgCl in pure water is 5.89×10^-6M. The correct choice is C.

If the Ksp for AgCl Is Equal to 4 AgCl2−\mathrm { AgCl } _ { 2 } ^ { - }AgCl2−​

If the K_sp for AgCl Is Equal to 4 $\mathrm { AgCl } _ { 2 } ^ { - }$