JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Equilibrium

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

2 5^{\circ} 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

]

Answer: (C)

Solution

The buffer of the weak base B with its conjugate acid

BH X^{+}

follows

pOH = p K_{b} + lo g \frac{[ BH X ^{+} ]}{[ B ]}

. At pH 9,

pOH = 5

, so

5 = 5.699 + lo g \frac{[ BH X ^{+} ]}{[ B ]}

, giving

lo g \frac{[ BH X ^{+} ]}{[ B ]} = - 0.699

and hence

\frac{[ BH X ^{+} ]}{[ B ]} = \frac{1}{5}

. The HCl added turns an equal amount of base into salt, so the salt is

0.02 x

mmol and the base left is

0.02 (y - x)

mmol, giving

\frac{x}{y - x} = \frac{1}{5}

and therefore

y = 6 x

. With the total volume

x + y = 100

mL,

7 x = 100

, so

x = 14.3

mL and

y = 85.7

mL.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option