Question

Calculate the [Zn²⁺] in a solution that is in equilibrium with ZnS(s) and in which [H₃O⁺] = 1.0x10^-5M and [H₂S] = 1.10M.

Short answer

The concentration of in the solution is $\boxed{2X{10}^{-13}M}$.

Step-by-step solution

Step(1):- The acid ionization

In addition, the acid ionization of H₂S must be considered:

$H_{2}S\left(aq\right)+H_{2}O\left(l\right)H_{3}O\left(aq\right)+H{S}^{-}\left(aq\right)$

The equilibrium expressions for this reaction gives

$\begin{array}{rcl}\frac{\left[H_2{O}^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}& =& K_{SP}\\ \frac{\left[1.0X10\right]\left[H{S}^{-}\right]}{\left[0.10\right]}& =& 9.1X{10}^{-8}\\ \left[H{S}^{-}\right]& =& \frac{9.1X{10}^{-8}X\left(0.10\right)}{\left(1X{10}^{-6}\right)}\\ \left[H{S}^{-}\right]& =& 9.1X{10}^{-4}M\end{array}$

Step(2):- The equation for the reaction

The equation for the reaction is

$ZnS\left(s\right)+H_{2}O\left(l\right)Z{n}^{2+}\left(aq\right)+H{S}^{-}\left(aq\right)+O{H}^{-}\left(aq\right)$

The equilibrium constant is

$$\begin{gathered} \left[Z{n}^{2+}\right]\left[H{S}^{-}\right]\left[O{H}^{-}\right]=2X{10}^{-25} \\ \left[Z{n}^{2+}\right]=\frac{2X{10}^{-25}}{\left[H{S}^{-}\right]\left[O{H}^{-}\right]} \\ \left[Z{n}^{2+}\right]=\frac{2X{10}^{-25}}{\left(9.1X{10}^{-4}\right)\left(1X{10}^{-9}\right)} \\ \left[Z{n}^{2+}\right]=2X{10}^{-23}M \end{gathered}$$

Therefore, the concentration of in the solution is $\boxed{2X{10}^{-13}M}$.