JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

Consider a weak base 'B' of pK $_{b}$ = 5.699. 'x' mL of 0.02 M HCl and 'y' mL of 0.02 M weak base 'B' are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of 'x' and 'y' respectively are:

[Given: $lo g 2 = 0.3010$ , $lo g 3 = 0.4771$ , $lo g 5 = 0.699$ ]

A
x y
11.1 88.9
B
x y
42.7 57.3
C
x y
14.3 85.7
D
x y
85.7 14.3

x	y
11.1	88.9

x	y
42.7	57.3

x	y
14.3	85.7

x	y
85.7	14.3

Solution

HCl is the limiting strong acid that converts part of B into its conjugate acid BH $^{+}$ . With $x$ mL of 0.02 M HCl and $y$ mL of 0.02 M B:

moles BH $^{+}$ formed $= 0.02 x$ (the salt), moles B left $= 0.02 (y - x)$ (the base).

pH = 9 $\Rightarrow$ pOH = 5. Henderson-Hasselbalch for a basic buffer:

$pOH = p K_{b} + lo g \frac{[ Salt ]}{[ Base ]}$

$5 = 5.699 + lo g \frac{x}{y - x}$

$lo g \frac{x}{y - x} = - 0.699 = lo g \frac{1}{5}$

$\Rightarrow \frac{x}{y - x} = \frac{1}{5} \Rightarrow 5 x = y - x \Rightarrow y = 6 x$ .

With $x + y = 100$ : $x + 6 x = 100 \Rightarrow x = \frac{100}{7} \approx 14.3$ mL, $y = \frac{600}{7} \approx 85.7$ mL.

$∴$ Option (3) is correct.

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