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JEE Advanced 2020 · Paper 2, Chemistry question 13

A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pKbK_b of the base? The neutralization reaction is given by B+HA→BHX++AX−\ce{B + HA -> BH+ + A-}.

pH against volume of HA added (0-12 mL) on a grid. The pH starts near 12, drops quickly to about 11 and then levels off near 10.5-10 up to about 6 mL, falls steeply at 6 mL (the equivalence point) to about 2.5, and then decreases slowly towards 1.5.
Answer and worked solution

Official answer: 2.8

Equivalence. The steep drop is at 6 mL, so half-neutralisation is at 3 mL. There [B]=[BH+][\mathrm B]=[\mathrm{BH^+}] and pOH=pKb\mathrm{pOH}=\mathrm pK_b.

Reading the curve. At 3 mL the pH is about 11, so pOH≈3\mathrm{pOH}\approx3 and pKb≈3\mathrm pK_b\approx3.

Check. The starting pH of 12 for 0.1 M base fits:
pOH=12(pKb−log⁡0.1)=12(3+1)=2.\mathrm{pOH}=\tfrac12(\mathrm pK_b-\log0.1)=\tfrac12(3+1)=2.
Answer: 3 (key: 2.80 to 3.20)

Topics:Indicators and titrationsBuffer solutions
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