Question

Solid potassium sulfide is slowly added to 150 mL of a copper(II) chloride solution until the concentration of sulfide ion is 0.0464 M. The maximum amount of copper(II) ion remaining in solution is ___ M

Answer 1

Answer:

$&=1.72 \times 10^{-35} \mathrm{M}$

Explanation:

The concentration of copper (II) ions is calculated by expression shown as,

$K_{s p}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{s}^{2-}\right]$

Here,

The solubility product of CuS is “ $K_{\text {sp }}$"

The concentration of copper ions is " $\left[\mathrm{Cu}^{2+}\right]$ "$. The concentration of sulfide ions is $"\left[\mathrm{~s}^{2-}\right]^{\prime \prime}$

The theoretical value of solubility product of $\mathrm{CuS}$ is $8 \times 10^{-37}$

Substitute the known values in equation (I).

$8 \times 10^{-37} &=\left[\mathrm{Cu}^{2+}\right] \times 0.0464$ $\left[\mathrm{Cu}^{2+}\right] &=\frac{8 \times 10^{-37}}{0.0464}$

$&=1.72 \times 10^{-35} \mathrm{M}$