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

Answer:

$5.5*10^{-12}M$

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:

$3.0\times10^{-23}=x^2$

Take the square root of both sides

$\begin{gathered} x=\sqrt{3.0\times10^{-23}} \\ x=5.47\times10^{-12}M \\ [Zn^{2+}]\approx5.5\times10^{-12}M \end{gathered}$

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