Question

Three solutions (A, B and C) are prepared according to given diagrams \[ \text{Solution A: } 0.1M,\;10\,\text{ml HCl(aq)} + 0.1M,\;25\,\text{ml Ca(OH)}_2\text{(aq)} \] \[ \text{Solution B: } 0.1M,\;10\,\text{ml H}_2\text{SO}_4\text{(aq)} + 0.1M,\;10\,\text{ml Ca(OH)}_2\text{(aq)} \] \[ \text{Solution C: } 0.1M,\;10\,\text{ml H}_2\text{SO}_4\text{(aq)} + 0.1M,\;10\,\text{ml NaOH(aq)} \] If pH of solutions A, B and C are respectively \(pH_1\), \(pH_2\) and \(pH_3\), then correct option will be:

• A. \(pH_3 < pH_2 < pH_1\)
• B. \(pH_3 > pH_2 > pH_1\)
• C. \(pH_3 > pH_1 > pH_2\)
• D. \(pH_1 < pH_3 < pH_2\)

Answer 1

The Correct Option is A

Concept:
pH comparison can be determined by calculating milli-equivalents of acids and bases. The nature of the resulting solution (acidic, basic, or neutral) depends on which component is in excess. Step 1: Solution A analysis. Milli-equivalent of HCl: \[ 0.1 \times 10 = 1 \] Milli-equivalent of \(Ca(OH)_2\): \[ 0.1 \times 25 \times 2 = 5 \] Since base \(>\) acid, the solution is basic. \[ pH_1 > 7 \] Step 2: Solution B analysis. Milli-equivalent of \(H_2SO_4\): \[ 0.1 \times 10 \times 2 = 2 \] Milli-equivalent of \(Ca(OH)_2\): \[ 0.1 \times 10 \times 2 = 2 \] Acid = Base, so the solution is neutral. \[ pH_2 = 7 \] Step 3: Solution C analysis. Milli-equivalent of \(H_2SO_4\): \[ 0.1 \times 10 \times 2 = 2 \] Milli-equivalent of NaOH: \[ 0.1 \times 10 = 1 \] Acid \(>\) Base, so the solution is acidic. \[ pH_3 < 7 \] Step 4: Final comparison. \[ pH_3 < pH_2 < pH_1 \]