Answer:
[tex]&=1.72 \times 10^{-35} \mathrm{M}[/tex]
Explanation:
The concentration of copper (II) ions is calculated by expression shown as,
[tex]K_{s p}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{s}^{2-}\right][/tex]
Here,
The solubility product of CuS is “ [tex]K_{\text {sp }}[/tex]"
The concentration of copper ions is " [tex]\left[\mathrm{Cu}^{2+}\right][/tex] "$. The concentration of sulfide ions is [tex]"\left[\mathrm{~s}^{2-}\right]^{\prime \prime}[/tex]
The theoretical value of solubility product of [tex]\mathrm{CuS}[/tex] is [tex]8 \times 10^{-37}[/tex]
Substitute the known values in equation (I).
[tex]8 \times 10^{-37} &=\left[\mathrm{Cu}^{2+}\right] \times 0.0464[/tex] [tex]\left[\mathrm{Cu}^{2+}\right] &=\frac{8 \times 10^{-37}}{0.0464}[/tex]
[tex]&=1.72 \times 10^{-35} \mathrm{M}[/tex]
