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ChemMedMain 2026 • 2nd Apr - Shift 2

Titration pH Calculation of Weak Monoprotic Acid

At 25∘C25^\circ\text{C}, 20.0 mL20.0\text{ mL} of 0.2 M0.2\text{ M} weak monoprotic acid HX\text{HX} is titrated against 0.2 M NaOH0.2\text{ M}\text{ NaOH}. The pH\text{pH} of the solution (a) at the start of the titration (when NaOH\text{NaOH} has not been added) and (b) when 10 mL10\text{ mL} of NaOH\text{NaOH} is added respectively, are :

Given : Ka=5×10−4K_a = 5 \times 10^{-4} pKa=3.3\text{p}K_a = 3.3 α≪1\alpha \ll 1

Options

A

(a) 0.7, (b) 2.0

B

(a) 2.0, (b) 3.3

Correct
C

(a) 1.1, (b) 2.2

D

(a) 3.0, (b) 2.2

Topics & Concepts

EquilibriumAcid-Base Equilibrium

Step-by-Step Solution

To find the pH\text{pH} of the solution at different stages of the titration, we analyze the system in two parts:

Part (a): At the start of the titration (0 mL of NaOH\text{NaOH} added)

At the beginning, the solution contains only the weak monoprotic acid HX\text{HX} with concentration C=0.2 MC = 0.2\text{ M} and dissociation constant Ka=5×10−4K_a = 5 \times 10^{-4}.

Since α≪1\alpha \ll 1, the degree of dissociation is small, and the concentration of hydrogen ions [H+][\text{H}^+] can be calculated using the approximation formula: [H+]=Ka⋅C[\text{H}^+] = \sqrt{K_a \cdot C}

Substitute the given values into the equation: [H+]=(5×10−4)×0.2=10−4=10−2 M[\text{H}^+] = \sqrt{(5 \times 10^{-4}) \times 0.2} = \sqrt{10^{-4}} = 10^{-2}\text{ M}

The pH\text{pH} at the start is: pH=−log⁡10([H+])=−log⁡10(10−2)=2.0\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(10^{-2}) = 2.0

Part (b): When 10 mL of 0.2 M NaOH\text{NaOH} is added

First, let's calculate the initial millimoles of the weak acid and the base:

  • Millimoles of HX=20.0 mL×0.2 M=4.0 mmol\text{Millimoles of HX} = 20.0\text{ mL} \times 0.2\text{ M} = 4.0\text{ mmol}
  • Millimoles of NaOH added=10.0 mL×0.2 M=2.0 mmol\text{Millimoles of NaOH added} = 10.0\text{ mL} \times 0.2\text{ M} = 2.0\text{ mmol}

The neutralization reaction occurs as follows: HX+NaOH→NaX+H2O\text{HX} + \text{NaOH} \rightarrow \text{NaX} + \text{H}_2\text{O}

Millimoles at equilibrium:

  • HX remaining=4.0 mmol−2.0 mmol=2.0 mmol\text{HX remaining} = 4.0\text{ mmol} - 2.0\text{ mmol} = 2.0\text{ mmol}
  • NaX formed=2.0 mmol\text{NaX formed} = 2.0\text{ mmol}

Since both the weak acid HX\text{HX} and its conjugate base X−\text{X}^- are present in equal amounts in the solution, an acidic buffer is formed. Applying the Henderson-Hasselbalch equation: pH=pKa+log⁡10([X−][HX])\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{X}^-]}{[\text{HX}]}\right)

pH=pKa+log⁡10(2.0 mmol2.0 mmol)=pKa+log⁡10(1)\text{pH} = \text{p}K_a + \log_{10}\left(\frac{2.0\text{ mmol}}{2.0\text{ mmol}}\right) = \text{p}K_a + \log_{10}(1)

pH=pKa\text{pH} = \text{p}K_a

Given that pKa=3.3\text{p}K_a = 3.3: pH=3.3\text{pH} = 3.3

Conclusion:

  • pH\text{pH} at the start =2.0= 2.0
  • pH\text{pH} after adding 10 mL10\text{ mL} of NaOH=3.3\text{NaOH} = 3.3

Therefore, the correct option is B.

Titration pH Calculation of Weak Monoprotic Acid | Chemistry PYQ Solution - JEE Challenger