Question

The solubility product constant (Ksp) for zinc sulfide is 3.0 x 10^-23 at 25 C. What is the concentration of zinc ions in a saturated zinc sulfide solution? Answer choices: 4.1 x 10^-3 M 5.5 x 10^-12 M 1.5 x 10^-23 M 1.5 x 10^-12 M

Answer 1

`5.5*10^(-12)M`

Explanations

The dissociation of zinc sulfide is expressed as:

`ZnS\rightarrow[Zn^(2+)][S^(2-)]`

The Ksp value is expressed as:

`\begin{gathered} Ksp=(x)(x) \\ Ksp=x^2 \end{gathered}`

Substitute the given Ksp value to have:

`3.0*10^(-23)=x^2`

Take the square root of both sides

`\begin{gathered} x=\sqrt{3.0*10^(-23)} \\ x=5.47*10^(-12)M \\ [Zn^(2+)]\approx5.5*10^(-12)M \end{gathered}`

Hence the concentration of zinc ions in a saturated zinc sulfide solution is 5.5 * 10^-12M