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Chemistry Question 56 – JEE-MAIN 2026

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At 25∘C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are : Given: Ka=5×10−4 pKa=3.3 α<<1

For a weak acid solution, the pH is determined by its dissociation constant and initial concentration. For a buffer solution, the pH depends on the pKa of the weak acid and the ratio of the concentrations of the conjugate base and weak acid.

● Video Walkthrough
Step 1: Calculate pH of initial weak acid solution (a)✦ Active

The initial solution contains only the weak acid HX. Using the approximation for weak acid dissociation ([H+]=Ka⋅CHX) since α<<1 is given:

[H+]=(5×10−4)×(0.2)=1×10−4=1×10−2 M

Therefore, the pH at the start of the titration is:

pH (a)=−log⁡(1×10−2)=2.0
Step 2: Calculate pH after adding NaOH (b)○ Expand

First, determine the initial moles of HX and the moles of NaOH added:

Initial moles of HX=20.0 mL×0.2 M=4.0 mmol Moles of NaOH added=10.0 mL×0.2 M=2.0 mmol

The reaction is HX + NaOH → NaX + H2O. After the reaction, 2.0 mmol of HX remains (4.0−2.0) and 2.0 mmol of NaX (conjugate base X−) is formed. This creates a buffer solution. Using the Henderson-Hasselbalch equation:

pH (b)=pKa+log⁡([X−][HX])

Since the moles of X− and HX are equal (2.0 mmol each), the ratio [X−][HX]=1. Given pKa=3.3:

pH (b)=3.3+log⁡(1)=3.3+0=3.3
💡 Teacher's Secret Hint

Remember that at the half-equivalence point of a weak acid-strong base titration, pH = pKa.

Step 3: Match calculated pH values with options○ Expand

The calculated pH values are (a) 2.0 and (b) 3.3. This corresponds to option "2".

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