Question

the solubility of ni₃(po₄)₂ in water at a particular temperature is 8.0 × 10⁻⁴ m. what is the kₛₚ for ni₃(po₄)₂?

Explanation:

Step1: Write the dissolution equation

$$\ce{Ni3(PO4)2(s) <=> 3Ni^{2 + }(aq) + 2PO^{3-}_{4}(aq)}$$
Let the solubility of $\ce{Ni3(PO4)2}$ be $s$. Given $s = 8.0\times10^{-4}\ M$.
From the stoichiometry of the dissolution equation, $[\ce{Ni^{2+}}]=3s$ and $[\ce{PO^{3 - }_{4}}]=2s$.

Step2: Write the expression for $K_{sp}$

The solubility - product constant expression for $\ce{Ni3(PO4)2}$ is $K_{sp}=[\ce{Ni^{2+}}]^{3}[\ce{PO^{3 - }_{4}}]^{2}$.
Substitute $[\ce{Ni^{2+}}]=3s$ and $[\ce{PO^{3 - }_{4}}]=2s$ into the $K_{sp}$ expression:
$$K_{sp}=(3s)^{3}(2s)^{2}$$
$$K_{sp}=27s^{3}\times4s^{2}$$
$$K_{sp}=108s^{5}$$

Step3: Substitute the value of $s$

Substitute $s = 8.0\times10^{-4}\ M$ into the $K_{sp}=108s^{5}$ equation.
$$s^{5}=(8.0\times 10^{-4})^{5}=8.0^{5}\times10^{-20}=32768\times10^{-20}=3.2768\times 10^{-16}$$
$$K_{sp}=108\times3.2768\times 10^{-16}$$
$$K_{sp}=3.539\times10^{-14}$$

Answer:

$3.5\times 10^{-14}$