Question

at a particular temperature, the solubility of in₂(so₄)₃ in water is 0.0045 m.
you have found ksp to be 2.0 × 10⁻¹⁰. if solid in₂(so₄)₃ is added to a solution that already contains 0.200 m na₂so₄, what will the new solubility of the solid be?

Explanation:

Step1: Write the dissolution equation

$$\text{In}_2(\text{SO}_4)_3(s) ightleftharpoons 2\text{In}^{3 +}(aq)+3\text{SO}_4^{2 -}(aq)$$

Let the solubility of $\text{In}_2(\text{SO}_4)_3$ be $s$. Then $[\text{In}^{3+}]=2s$ and $[\text{SO}_4^{2 -}] = 3s+0.200$ (since $\text{Na}_2\text{SO}_4$ provides $0.200\ M$ of $\text{SO}_4^{2 -}$ initially).

Step2: Write the $K_{sp}$ expression

$$K_{sp}=[\text{In}^{3 +}]^2[\text{SO}_4^{2 -}]^3$$
Substitute $[\text{In}^{3+}]=2s$ and $[\text{SO}_4^{2 -}] = 3s + 0.200$ into the $K_{sp}$ expression. Given $K_{sp}=2.0\times 10^{-10}$
$$2.0\times 10^{-10}=(2s)^2(3s + 0.200)^3$$
Since $s$ is very small (common - ion effect), $3s+0.200\approx0.200$

Step3: Simplify the equation

$$2.0\times 10^{-10}=(4s^2)(0.200)^3$$
First, calculate $(0.200)^3=0.008$
The equation becomes $2.0\times 10^{-10}=4s^2\times0.008$
$$4s^2=\frac{2.0\times 10^{-10}}{0.008}=2.5\times 10^{-8}$$
$$s^2=\frac{2.5\times 10^{-8}}{4}=6.25\times 10^{-9}$$

Step4: Solve for $s$

$$s=\sqrt{6.25\times 10^{-9}} = 7.9\times 10^{-5}\ M$$

Answer:

The new solubility of $\text{In}_2(\text{SO}_4)_3$ is $7.9\times 10^{-5}\ M$