1. Interpret the given data

• $100\,\text{m mol}$ of $Ca(OH)_2 = 0.100\,\text{mol}$
• Mass of $Na_2SO_4 = 2\,\text{g}$

So, $n(Na_2SO_4)=\frac{2}{143}=0.013986\approx 0.014\,\text{mol}$

Volume of solution $=100\,\text{mL}=0.1\,\text{L}$.

2. Reaction between calcium hydroxide and sodium sulphate

The reaction is: $Ca(OH)_2 + Na_2SO_4 \rightarrow CaSO_4\downarrow + 2NaOH$

This is a $1:1$ reaction between $Ca(OH)_2$ and $Na_2SO_4$.

Since $n(Ca(OH)_2)=0.100\,\text{mol},\qquad n(Na_2SO_4)=0.014\,\text{mol}$

$Na_2SO_4$ is the limiting reagent.

Hence moles of $CaSO_4$ formed initially: $n(CaSO_4)=0.014\,\text{mol}$

Mass of $CaSO_4$ formed: $m=0.014\times 136=1.904\,\text{g}$

So, $m(CaSO_4)\approx 1.9\,\text{g}$

3. Amount of $Ca(OH)_2$ left after precipitation reaction

Consumed $Ca(OH)_2 = 0.014\,\text{mol}$

Leftover: $0.100-0.014=0.086\,\text{mol}$

In $0.1\,\text{L}$, if this remained dissolved, its concentration would be $C=\frac{0.086}{0.1}=0.86\,\text{M}$

But $Ca(OH)_2$ is sparingly soluble, so its dissolved concentration is controlled by $K_{sp}$.

4. Use solubility product of $Ca(OH)_2$

Dissolution: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+} + 2OH^-$

Let solubility of dissolved $Ca(OH)_2$ be $s$ mol L$^{-1}$. Then, $[Ca^{2+}] = s, \qquad [OH^-] = 2s$

Given: $K_{sp}= [Ca^{2+}][OH^-]^2 = s(2s)^2 = 4s^3 = 5.5\times 10^{-6}$

So, $s^3=\frac{5.5\times 10^{-6}}{4}=1.375\times 10^{-6}$

$s\approx 0.0111\,\text{M}$

Therefore, $[OH^-]=2s\approx 2(0.0111)=0.0222\,\text{M}$

This is the hydroxide concentration contributed by saturated $Ca(OH)_2$ alone.

5. But reaction produced $NaOH$ also

From the precipitation reaction, $Ca(OH)_2 + Na_2SO_4 \rightarrow CaSO_4 + 2NaOH$

If $0.014$ mol $Na_2SO_4$ reacts, then moles of $NaOH$ formed: $2\times 0.014=0.028\,\text{mol}$

In $0.1\,\text{L}$, $[NaOH]=\frac{0.028}{0.1}=0.28\,\text{M}$

Since $NaOH$ is a strong electrolyte, it gives $[OH^-]=0.28\,\text{M}$

This common ion suppresses the solubility of $Ca(OH)_2$ even further, so the dominant hydroxide concentration is from $NaOH$.

Thus, $[OH^-]\approx 0.28\,\text{mol L}^{-1}$

6. Final answer

• Mass of $CaSO_4$ formed $\approx 1.9\,\text{g}$
• Concentration of $OH^- \approx 0.28\,\text{mol L}^{-1}$

Therefore, the correct option is: $\boxed{\text{C}}$

7. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.