JEE Main 2026 · 4 Apr · Shift 1 · Chemistry Q74

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ChemistryInteger+4 / -1Question 74 of 75

The pH of a solution obtained by mixing 5 mL of 0.1 M $\ce{NH4OH}$ solution with 250 mL of 0.1 M $\ce{NH4Cl}$ solution is ________ $\times10^{-2}$ . (Nearest integer)
Given: $pK_b(\ce{NH4OH})=4.74$ , $\log2=0.30$ , $\log3=0.48$ , $\log5=0.70$

Your answer

Answer and worked solution

Official answer: 756

Basic buffer. $\ce{NH4OH}$ 0.5 mmol and $\ce{NH4Cl}$ 25 mmol:
$\text{pOH}=4.74+\log\frac{25}{0.5}=4.74+1.70=6.44.$

pH $=7.56=756\times10^{-2}$ .

Answer: 756

Topics:Buffer solutions

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JEE Main 2026 · 4 Apr · Shift 1, Chemistry question 74