JEE Chemistry 2023 Question with Solution

Given: $\text{pH} = 12$, volume of solution $= 100 \, \text{mL}$. Calcium hydroxide dissociates completely.

Find: The value of $x$ in $x \times 10^{-1}$ millimoles of $\text{Ca(OH)}_2$.

Using the relation:

$$\text{pH} + \text{pOH} = 14$$

we get:

$$\text{pOH} = 14 - 12 = 2$$

Hence, hydroxide ion concentration is:

$$[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{M}$$

Calcium hydroxide dissociates as:

$$\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$$

So, from stoichiometry:

$$[\text{Ca(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{M}$$

Millimoles of $\text{Ca(OH)}_2$ in $100 \, \text{mL}$ are:

$$\text{millimoles} = \text{Molarity} \times \text{Volume (in mL)} = 5 \times 10^{-3} \times 100 = 5 \times 10^{-1}$$

Comparing with $x \times 10^{-1}$, we get:

$$x = 5$$

Therefore, the required value is $5$.

Given: $\text{pH} = 12$ and solution volume $100 \, \text{mL}$.

Find: Millimoles of $\text{Ca(OH)}_2$ expressed as $x \times 10^{-1}$.

1. Convert pH to pOH.

$$\text{pOH} = 14 - 12 = 2$$

2. Use pOH to calculate hydroxide concentration.

$$[\text{OH}^-] = 10^{-2} \, \text{mol L}^{-1}$$

3. One mole of $\text{Ca(OH)}_2$ gives $2$ moles of $\text{OH}^-$.

$$\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$$

Therefore:

$$[\text{Ca(OH)}_2] = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{mol L}^{-1}$$

4. Convert concentration to millimoles in $100 \, \text{mL}$.

$$\text{millimoles} = 5 \times 10^{-3} \times 100 = 5 \times 10^{-1}$$

Thus, in $x \times 10^{-1}$, the value of $x$ is $5$.

Therefore, the answer is $5$.