Answer to Question #98505 in Physical Chemistry for kiv

Question

In 1N HCl solution, we have 0.1M Cu++ and Mn++ ions. To this solution we add dihydrogen sulfide gas which has a solubility of 0.1 in 1N HCl. If the equilibrium constant for dihydrogensulfide is 10^(-22), then
1)The concentration of sulfide ions is 10^(-23)
2)CuS will be precipitated but MnS will not
3)Both CuS and MnS will be precipitated
4)Neither CuS nor MnS will precipitate out

Answer 1

Solution.

$C(H^+) = C(HCl)$

$K1 \times K2 = 10^{-22}$

$K = \frac{[H^+]^2 \times [S^{2-}]}{[H2S]}$

$[S^{2-}] = \frac{K \times [H2S]}{[H^+]^2}$

$[S^{2-}] = 10^{-23}$

$T(CuS) = [Cu^{2+}] \times [S^{2-}]$

$T(CuS) = 0.1 \times 10^{-23}$

$T(MnS) = [Mn^{2+}] \times [S^{2-}]$

$T(MnS) = 0.1 \times 10^{-23}$

Since the solubility product (theoretical) is less than the value we found, the precipitate will only form CuS, and MnS will not precipitate.

Answer:

1), 2)