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 (CaSO₄) formed and the concentration of hydroxide ions (OH⁻) in the resulting solution after mixing 100 mmol of Ca(OH)₂ and 2 g of sodium sulfate (Na₂SO₄) in water.

### Step 1: Calculate moles of sodium sulfate (Na₂SO₄)
Given:
- Mass of Na₂SO₄ = 2 g
- Molar mass of Na₂SO₄ = 143 g/mol

To find the moles of Na₂SO₄:
\[ \text{Moles of Na₂SO₄} = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \text{ g}}{143 \text{ g/mol}} \approx 0.01399 \text{ mol} \approx 14 \text{ mmol} \]

### Step 2: Write the balanced chemical equation
The reaction between Ca(OH)₂ and Na₂SO₄ can be represented as:
\[ \text{Ca(OH)}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{NaOH} \]

### Step 3: Determine the limiting reactant
From the balanced equation:
- 1 mole of Ca(OH)₂ reacts with 1 mole of Na₂SO₄ to produce 1 mole of CaSO₄.

We have:
- 100 mmol of Ca(OH)₂
- 14 mmol of Na₂SO₄

Since Na₂SO₄ is the limiting reactant (14 mmol < 100 mmol), it will determine the amount of CaSO₄ produced.

### Step 4: Calculate the moles of CaSO₄ produced
From the stoichiometry of the reaction:
- 1 mole of Na₂SO₄ produces 1 mole of CaSO₄.

Thus, 14 mmol of Na₂SO₄ will produce 14 mmol of CaSO₄.

### Step 5: Calculate the mass of CaSO₄ produced
Molar mass of CaSO₄ = 136 g/mol
\[ \text{Mass of CaSO₄} = \text{moles} \times \text{molar mass} = 0.014 \text{ mol} \times 136 \text{ g/mol} = 1.904 \text{ g} \approx 1.9 \text{ g} \]

### Step 6: Calculate the concentration of OH⁻ ions
From the reaction, 2 moles of NaOH are produced for every mole of Na₂SO₄ reacted:
\[ \text{Moles of NaOH} = 2 \times \text{moles of Na₂SO₄} = 2 \times 0.014 \text{ mol} = 0.028 \text{ mol} = 28 \text{ mmol} \]

Total volume of the solution = 100 mL = 0.1 L

Concentration of OH⁻:
\[ \text{Concentration of OH}^- = \frac{\text{moles of OH}^-}{\text{volume in L}} = \frac{0.028 \text{ mol}}{0.1 \text{ L}} = 0.28 \text{ M} \]

### Final Answers
- Mass of CaSO₄ formed: **1.9 g**
- Concentration of OH⁻: **0.28 M**