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

`&=1.72 * 10^(-35) \mathrm{M}`

Step-by-step 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

The theoretical value of solubility product of
`\mathrm{CuS}` is
`8 * 10^(-37)`

Substitute the known values in equation (I).

`8 * 10^(-37) &=\left[\mathrm{Cu}^(2+)\right] * 0.0464`
`\left[\mathrm{Cu}^(2+)\right] &=(8 * 10^(-37))/(0.0464)`

`&=1.72 * 10^(-35) \mathrm{M}`