Question

Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is \(x \times 10^{-1}\). The value of \(x\) is — (Nearest integer).

Hint:

For pH-based calculations:
• Use the relationship pH + pOH = 14 to find OH− concentration.
• Consider the stoichiometry of the dissociation reaction to relate hydroxide
ion concentration to the base concentration.
• Calculate millimoles using the formula Molarity × Volume (in mL).

Answer 1

Correct Answer: 5

1.Given pH: The pH of the solution is given as 12. From the relation:
\[\text{pH} + \text{pOH} = 14,\]
we find:
\[\text{pOH} = 14 - 12 = 2.\]
2.Hydroxide Ion Concentration: The concentration of OH\(^-\) ions is:
\[[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2}~\text{M}.\]
3. Calcium Hydroxide Dissociation: Calcium hydroxide dissociates completely as:
\[\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-.\]
From stoichiometry, the concentration of \(\text{Ca(OH)}_2\) is half of the OH\(^-\) concentration:
\[[\text{Ca(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{10^{-2}}{2} = 5 \times 10^{-3}~\text{M}.\]
4. Millimoles of \(\text{Ca(OH)}_2\): The number of millimoles of \(\text{Ca(OH)}_2\) in 100 mL of solution is:
\[\text{Millimoles of } \text{Ca(OH)}_2 = \text{Molarity} \times \text{Volume (in mL)} = 5 \times 10^{-3} \times 100 = 5 \times 10^{-1}.\]
5. Value of \(x\): Comparing with \(x \times 10^{-1}\), we find:
\[x = 5.\]