The pH a solution obtained by mixing 5 mL of $0.1 \mathrm{M}$ $\mathrm{NH}_{4}\mathrm{OH}$ solution with 250 mL of 0.1 M $\mathrm{NH}_{4}\mathrm{Cl}$ solution is ______ $\times 10^{-2}$. (Nearest integer) Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{4}\mathrm{OH}\right)=4.74$ $$\begin{aligned} & \log 2=0.30 \\ & \log 3=0.48 \\ & \log 5=0.70 \end{aligned}$$

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Equilibrium

The pH a solution obtained by mixing 5 mL of

0.1 M

NH_{4} OH

solution with 250 mL of 0.1 M

NH_{4} Cl

solution is ______

\times 1 0^{- 2}

. (Nearest integer)

Given :

pK_{b} (NH_{4} OH) = 4.74

lo g 2 = 0.30 lo g 3 = 0.48 lo g 5 = 0.70

Answer: 756

The mixture is a basic buffer of

NH X_{4} OH

and its salt

NH X_{4} Cl

. The amounts present are

NH X_{4} OH = 5 \times 0.1 = 0.5

mmol and

NH X_{4} Cl = 250 \times 0.1 = 25

mmol. For a basic buffer

pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]} = 4.74 + lo g \frac{25}{0.5} = 4.74 + lo g 50

. Since

lo g 50 = 2 - lo g 2 = 2 - 0.30 = 1.70

pOH = 6.44

and

pH = 14 - 6.44 = 7.56

. Expressed as asked, the pH is

756 \times 1 0^{- 2}

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer