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 0.150 m in(no₃)₃, what will the new solubility of the solid be?

Explanation:

Step1: Write the dissolution equation

$$\ce{In2(SO4)3(s)<=>2In^{3 + }(aq) + 3SO^{2 - }_4(aq)}$$
Let the solubility of $\ce{In2(SO4)3}$ be $s$. The concentration of $\ce{In^{3+}}$ from $\ce{In(NO3)3}$ is $0.150\ M$, and from the dissolution of $\ce{In2(SO4)3}$ is $2s$. So the total $[\ce{In^{3+}}]=0.150 + 2s$. The concentration of $\ce{SO^{2-}_4}$ from the dissolution of $\ce{In2(SO4)3}$ is $3s$.

Step2: Write the $K_{sp}$ expression

$$K_{sp}=[\ce{In^{3+}}]^2[\ce{SO^{2-}_4}]^3$$
Substitute $[\ce{In^{3+}}]=0.150 + 2s$ and $[\ce{SO^{2-}_4}]=3s$ into the $K_{sp}$ expression. Since $K_{sp}=2.0\times 10^{-10}$ and $s$ is very small (common - ion effect), we can approximate $0.150+2s\approx0.150$.
So $K_{sp}=(0.150)^2\times(3s)^3$

Step3: Solve for $s$

$$2.0\times 10^{-10}=(0.150)^2\times27s^3$$
First, calculate $(0.150)^2 = 0.0225$. Then the equation becomes:
$$2.0\times 10^{-10}=0.0225\times27s^3$$
$$2.0\times 10^{-10}=0.6075s^3$$
$$s^3=\frac{2.0\times 10^{-10}}{0.6075}\approx3.29\times 10^{-10}$$
$$s=\sqrt[3]{3.29\times 10^{-10}}\approx6.9\times 10^{-4}\ M$$

Answer:

The new solubility of $\ce{In2(SO4)3}$ is approximately $6.9\times 10^{-4}\ M$.