A mixture of 100m mol of `Ca(OH)_(2)` and 2g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of `OH^(-)` in resulting solution, respectively, are: (Molar mass of `Ca(OH)_(2),Na_(2)SO_(4)` and `CaSO_(4)` are 74, 143 and 136 g `mol^(-1)` respectively, `K_(sp)` of `Ca(OH)_(2)` is `5.5xx10^(-6)` )

Options

A. `13.6g,0.28molL^(-1)`
B. `1.9g,0.14molL^(-1)`
C. `1.9g,0.28molL^(-1)`
D. `13.6g,0.14molL^(-1)`

Explanation

To solve the problem, we need to determine the mass of calcium sulfate formed and the concentration of hydroxide ions in the resulting solution after mixing the given amounts of calcium hydroxide and sodium sulfate.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between sodium sulfate (Na₂SO₄) and calcium hydroxide (Ca(OH)₂) can be represented as:
\[ \text{Na}_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2\text{NaOH} \]

2. **Calculate Moles of Sodium Sulfate**:
Given the mass of sodium sulfate is 2 g, we can calculate the number of moles using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of Na₂SO₄ is 143 g/mol. Therefore:
\[ \text{Moles of Na}_2\text{SO}_4 = \frac{2 \, \text{g}}{143 \, \text{g/mol}} \approx 0.01398 \, \text{mol} \, \text{or} \, 13.98 \, \text{mmol} \]

3. **Determine Moles of Calcium Hydroxide**:
We are given that there are 100 mmol of Ca(OH)₂ available.

4. **Identify the Limiting Reagent**:
From the balanced equation, 1 mole of Na₂SO₄ reacts with 1 mole of Ca(OH)₂. Since we have 13.98 mmol of Na₂SO₄ and 100 mmol of Ca(OH)₂, Na₂SO₄ is the limiting reagent.

5. **Calculate Moles of Calcium Sulfate Formed**:
According to the stoichiometry of the reaction, the moles of CaSO₄ formed will be equal to the moles of Na₂SO₄ reacted:
\[ \text{Moles of CaSO}_4 = \text{Moles of Na}_2\text{SO}_4 = 13.98 \, \text{mmol} \]

6. **Calculate Mass of Calcium Sulfate Formed**:
Using the molar mass of CaSO₄ (136 g/mol), we can find the mass:
\[ \text{Mass of CaSO}_4 = \text{Moles} \times \text{Molar Mass} = 13.98 \times 10^{-3} \, \text{mol} \times 136 \, \text{g/mol} \approx 1.90 \, \text{g} \]

7. **Calculate Moles of Hydroxide Ions**:
From the balanced equation, for every mole of Na₂SO₄, 2 moles of NaOH are produced. Therefore:
\[ \text{Moles of NaOH} = 2 \times \text{Moles of Na}_2\text{SO}_4 = 2 \times 13.98 \, \text{mmol} = 27.96 \, \text{mmol} \]

8. **Calculate Concentration of Hydroxide Ions**:
The total volume of the solution is 100 mL (0.1 L). The concentration of OH⁻ ions can be calculated as:
\[ \text{Concentration of OH}^- = \frac{\text{Moles of NaOH}}{\text{Volume in L}} = \frac{27.96 \times 10^{-3} \, \text{mol}}{0.1 \, \text{L}} = 0.2796 \, \text{mol/L} \approx 0.28 \, \text{mol/L} \]

### Final Answers:
- Mass of CaSO₄ formed: **1.90 g**
- Concentration of OH⁻ in the resulting solution: **0.28 mol/L**