Question
The $K_{s p}$ values of two slightly soluble salts $A B$ and $\mathrm{PQ}_2$ are each equal to $4.0 \times 10^{-18}$. Which salt is more soluble?
✓
Answer
For $A B$ salt, $K_{s p}=\left[A^{+}\right]\left[B^{-}\right]$If $x$ is the solubolity, then
$x^2=4.0 \times 10^{-18}$
$\text { or } x=2.0 \times 10^{-9}$
For $\mathrm{PQ}_2$ salt, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{P}^{2+}\right]\left[\mathrm{Q}^{-}\right]^2$
If y is the solubility, then
$4 y^3=K_{\text {sp }}=4.0 \times 10^{-18}$
or $y=1.0 \times 10^{-6}$
Since $y$ is more than $x, P Q_2$ salt has more solubility.
$x^2=4.0 \times 10^{-18}$
$\text { or } x=2.0 \times 10^{-9}$
For $\mathrm{PQ}_2$ salt, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{P}^{2+}\right]\left[\mathrm{Q}^{-}\right]^2$
If y is the solubility, then
$4 y^3=K_{\text {sp }}=4.0 \times 10^{-18}$
or $y=1.0 \times 10^{-6}$
Since $y$ is more than $x, P Q_2$ salt has more solubility.
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