Question

Calculate the$\mathbf{\left[}\mathbf{C}{\mathbf{d}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}$ in a solution that is in equilibrium with $\mathbf{Cds}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$and in which $\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{M}$and.$\mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\mathbf{\right]}\mathbf{=}\mathbf{0}\mathbf{.10}\mathit{M}$

Short answer

The concentration of$\left[C{d}^{2+}\right]$ ion in a solution is .$\boxed{8X{10}^{-12}M}$

Step-by-step solution

Step-(1):-Finding the concentration of

$\begin{array}{l}pOH=14-pH\\ =14-3\\ =11\end{array}$We have the concentration of following they are$\left[H_{2}O\right]=1.0X{10}^{-3}M$ and $\left[H_{2}S\right]=0.01M$respectively.

Since, the pH of hydronium ion is given by:

$\begin{array}{l}=-\log \left[H_3{O}^{+}\right]\\ =-\log \left[1.0X{10}^{-3}\right]\\ =3\end{array}$

Similarly, the$pOH$of the solution is given by,

$\begin{array}{l}pH+pOH=14\\ \end{array}$

So, the concentration of .$\left[O{H}^{-}\right]=1.0X{10}^{-11}$

Step(2):- Finding the concentration of[Cd2+][Cd2+]

In addition, the acid ionization of $H_{2}S$ must be considered:

$H_{2}S\left(aq\right)+H_{2}O\left(aq\right)$$H_{{}_3}O\left(aq\right)+H{S}^{-}\left(aq\right)$

The equilibrium expression for this reaction gives

$\begin{array}{l}\frac{\left[H_{3}O\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}=K_{SP}\\ \frac{\left[1X{10}^{-3}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}=9.1X{10}^{-8}[\sin ce,K_{SP}=9.1X{10}^{-8}]\\ \left[H{S}^{-}\right]=\frac{9.1X{10}^{-8}X\left(0.10\right)}{\left(1X{10}^{-3}\right)}\\ =9.1X{10}^{-6}M\end{array}$

The equation for the reaction:-

$CdS\left(s\right)+H_{2}O\left(l\right)$

$\begin{array}{l}C{d}^{2+}\left(aq\right)+H{S}^{-}\left(aq\right)+O{H}^{-}\left(aq\right)\\ \end{array}$

The equilibrium constant is

$\begin{array}{l}\left[C{d}^{2+}\right]\left[H{S}^{-}\right]\left[O{H}^{-}\right]=7X{10}^{-28}\\ \left[C{d}^{2+}\right]=\frac{7X{10}^{-28}}{\left[H{S}^{-}\right]\left[O{H}^{-}\right]}\\ \left[C{d}^{2+}\right]=\frac{2X{10}^{-25}}{\left(9.1X{10}^{-6}\right)\left(1X{10}^{-11}\right)}\\ \left[C{d}^{2+}\right]=8X{10}^{-12}M\end{array}$

Therefore, the concentration of$\left[C{d}^{2+}\right]$ ion in a solution is .$\boxed{8X{10}^{-12}M}$