Arrange Resultant Acid Base Mixtures in Order of pH

To determine the increasing order of pH for the given resultant mixtures, we need to calculate the concentration of excess $\text{H}^+$ or $\text{OH}^-$ ions in each case.

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Analysis of Mixture A:

$10\text{ mL } 0.2\text{ M Ca(OH)}_2 + 25\text{ mL } 0.1\text{ M HCl}$

1. Milliequivalents of $\text{OH}^-$ ions from $\text{Ca(OH)}_2$: $\text{meq of OH}^- = \text{Volume (mL)} \times \text{Molarity} \times n\text{-factor}$ $\text{meq of OH}^- = 10 \times 0.2 \times 2 = 4.0\text{ meq}$

2. Milliequivalents of $\text{H}^+$ ions from $\text{HCl}$: $\text{meq of H}^+ = 25 \times 0.1 \times 1 = 2.5\text{ meq}$

3. Excess ions: $\text{Excess meq of OH}^- = 4.0 - 2.5 = 1.5\text{ meq}$

4. Concentration of $\text{OH}^-$ ions: $\text{Total volume } V = 10 + 25 = 35\text{ mL}$ $[\text{OH}^-] = \frac{1.5\text{ meq}}{35\text{ mL}} = \frac{3}{70}\text{ M} \approx 0.0428\text{ M}$

Since $\text{OH}^-$ is in excess, the solution is basic, giving a high $\text{pH}$ value ($\text{pH} > 7$). $\text{pOH} = -\log_{10}(0.0428) \approx 1.37 \implies \text{pH} = 14 - 1.37 = 12.63$

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Analysis of Mixture B:

$10\text{ mL } 0.01\text{ M H}_2\text{SO}_4 + 10\text{ mL } 0.01\text{ M Ca(OH)}_2$

1. Milliequivalents of $\text{H}^+$ ions from $\text{H}_2\text{SO}_4$: $\text{meq of H}^+ = 10 \times 0.01 \times 2 = 0.2\text{ meq}$

2. Milliequivalents of $\text{OH}^-$ ions from $\text{Ca(OH)}_2$: $\text{meq of OH}^- = 10 \times 0.01 \times 2 = 0.2\text{ meq}$

3. Resulting Solution: $\text{meq of H}^+ = \text{meq of OH}^- = 0.2\text{ meq}$

Since complete neutralization occurs between a strong acid and a strong base, the resulting salt ($\text{CaSO}_4$) does not undergo significant hydrolysis. Thus, the solution is neutral. $\text{pH} = 7.0$

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Analysis of Mixture C:

$10\text{ mL } 0.1\text{ M H}_2\text{SO}_4 + 10\text{ mL } 0.1\text{ M KOH}$

1. Milliequivalents of $\text{H}^+$ ions from $\text{H}_2\text{SO}_4$: $\text{meq of H}^+ = 10 \times 0.1 \times 2 = 2.0\text{ meq}$

2. Milliequivalents of $\text{OH}^-$ ions from $\text{KOH}$: $\text{meq of OH}^- = 10 \times 0.1 \times 1 = 1.0\text{ meq}$

3. Excess ions: $\text{Excess meq of H}^+ = 2.0 - 1.0 = 1.0\text{ meq}$

4. Concentration of $\text{H}^+$ ions: $\text{Total volume } V = 10 + 10 = 20\text{ mL}$ $[\text{H}^+] = \frac{1.0\text{ meq}}{20\text{ mL}} = 0.05\text{ M}$

5. pH calculation: $\text{pH} = -\log_{10}(0.05) = -\log_{10}\left(\frac{5}{100}\right) = 2 - \log_{10}(5) \approx 2 - 0.699 = 1.301$

Since $\text{H}^+$ is in excess, the solution is acidic ($\text{pH} < 7$).

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Conclusion:

Comparing the pH values of the three mixtures:

• Mixture C: $\text{pH} \approx 1.301$ (Acidic)
• Mixture B: $\text{pH} = 7.0$ (Neutral)
• Mixture A: $\text{pH} \approx 12.63$ (Basic)

Thus, the increasing order of pH values is: $\text{C < B < A}$

This corresponds to Option C.