Exercise 9 | Solubility and Precipitation Reactions

Chemical reaction: CuS (s) $\rightleftarrows$ Cu²⁺ (aq) + S^2- (aq)

K_sp = [Cu²⁺][S^2-]

The only source of Cu²⁺ is CuS ⇒ the total solubility s of CuS (s) in M is: s = [Cu²⁺]

⇒ s = $\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left[{\mathrm{S}}^{2-}\right]}$

Let's determine [S^2-]:

H₂S + H₂O $\rightleftarrows$ HS^- + H₃O⁺   [K_a]

K_a = $\frac{\left[{\mathrm{HS}}^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}$

HS^- + H₂O $\rightleftarrows$ S^2- + H₃O⁺   [K_a']

K_a’ = $\frac{\left[{\mathrm{S}}^{2-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{HS}}^{-}\right]}$

[S^2-] = K_a’ $\frac{\left[{\mathrm{HS}}^{-}\right]}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}$ = K_a’ K_a $\frac{\left[{\mathrm{H}}_2\mathrm{S}\right]}{{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2}$

pH = - log [H₃O⁺]

⇒ [H₃O⁺] = 1.00 x 10^-4 M

⇒ [S^2-] = 2.14 x 10^-13 M

⇒ s = K_sp / [S^2-] = 2.95 x 10^-23 M