Verified Solution • 2025–26 Session moderate

QJEEM23Chf4e0: JEE Main 2023 Chemistry - Class 11 JEEM23Chf4e0 Solution: A litre of buffer solution contains 0.1 mole of each of NH and…

A litre of buffer solution contains 0.1 mole of each of NH 3_3 and NH 4_4 Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____________ ×\times 10 −3^{-3} (Nearest integer)

[Given : pKb(NH3)=4.745\mathrm{pK_b(NH_3)=4.745}

log⁡2=0.301\mathrm{\log2=0.301}

log⁡3=0.477\mathrm{\log3=0.477}

T=298 K]\mathrm{T=298~K]}

Step-by-Step Verified Solution

In resultant solution

amp;nNH3=0.1−0.02=0.08amp;nNH4Cl=nNH4+=0.1+0.02=0.12amp;pOH=pKb+log⁡[NH4+][NH3]amp;=4.745+log⁡0.120.08amp;=4.745+log⁡32amp;=4.745+0.477−0.301amp;pOH=4.921amp;pH=14−pHamp;=9.079=9079×10−3\begin{aligned}& \mathrm{n}_{\mathrm{NH}_3}= 0.1-0.02=0.08 \\\\& \mathrm{n}_{\mathrm{NH}_4 \mathrm{Cl}}= \mathrm{n}_{\mathrm{NH}_4^{+}}=0.1+0.02=0.12 \\\\& \mathrm{pOH}= \mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \\\\&=4.745+\log \frac{0.12}{0.08} \\\\&=4.745+\log \frac{3}{2} \\\\&=4.745+0.477-0.301 \\\\& \mathrm{pOH}=4.921 \\\\& \mathrm{pH}=14-\mathrm{pH} \\\\&=9.079 = 9079\times 10^{-3}\end{aligned}

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