# Exercise 9 | Solubility and Precipitation Reactions

Calculate the solubility of CuS (s) in an aqueous solution buffered at pH = 4.00 and saturated with H2S so that [H2S] = 0.200 M.

Data: Ksp (CuS) = 6.30 x 10-36 2

Ka (H2S/HS-) = 8.90 x 10-8 M

Ka’ (HS-/S2-) = 1.20 x 10-13 M

Chemical reaction: CuS (s) $⇄$ Cu2+ (aq) + S2- (aq)

Ksp = [Cu2+][S2-]

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 [S2-]:

H2S + H2O $⇄$ HS- + H3O+   [Ka]

Ka = $\frac{\left[{\mathrm{HS}}^{-}\right]\left[{\mathrm{H}}_{3}{\mathrm{O}}^{+}\right]}{\left[{\mathrm{H}}_{2}\mathrm{S}\right]}$

HS- + H2O $⇄$ S2- + H3O+   [Ka']

Ka’ = $\frac{\left[{\mathrm{S}}^{2-}\right]\left[{\mathrm{H}}_{3}{\mathrm{O}}^{+}\right]}{\left[{\mathrm{HS}}^{-}\right]}$

[S2-] = Ka$\frac{\left[{\mathrm{HS}}^{-}\right]}{\left[{\mathrm{H}}_{3}{\mathrm{O}}^{+}\right]}$ = Ka’ Ka $\frac{\left[{\mathrm{H}}_{2}\mathrm{S}\right]}{{\left[{\mathrm{H}}_{3}{\mathrm{O}}^{+}\right]}^{2}}$

pH = - log [H3O+]

⇒ [H3O+] = 1.00 x 10-4 M

[S2-] = 2.14 x 10-13 M

s = Ksp / [S2-] = 2.95 x 10-23 M