JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Equilibrium

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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

p K_{b}

of the base? The neutralization reaction is given by

B + HA \to BH^{+} + A^{-}

Answer: 3

Solution

From the titration curve, the equivalence point occurs at V = 6 mL where pH ≈ 6 to 6.28.

At the equivalence point, all base B is converted to its conjugate acid BH⁺:

[BH^{+}] = \frac{0.1 \times V}{2 V} = 0.05 M

For a weak acid (BH⁺):

pH = 7 - \frac{1}{2} [pK_{b} + lo g (0.05)]

Using pH ≈ 6:

6 = 7 - \frac{1}{2} [pK_{b} - 1.3]

pK_{b} \approx 3.0

Alternatively, at half-equivalence point (V = 3 mL), pH = 11:

pOH = 14 - 11 = 3 = pK_{b}

Therefore,

pK_{b} = 3.0

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer

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