JEE Main 2017 · 9 April · Online · Chemistry · Ionic Equilibrium
B. 9.259 . 259.25
NH3+HCl⇌NH4Clmolesatt=050×.2×10−325×.2×10−3−molesatt=t25×.2×10−3025×.2×10−3\begin{array}{ll}{NH}_{3} + HCl ⇌ {NH}_{4}Cl \\ molesat\text{}t = 050 \times . 2 \times {10}^{- 3}25 \times . 2 \times {10}^{- 3} - \\ molesat\text{}t = t25 \times . 2 \times {10}^{- 3}025 \times . 2 \times {10}^{- 3}\end{array}NH3+HCl⇌NH4Clmolesatt=050×.2×10−325×.2×10−3−molesatt=t25×.2×10−3025×.2×10−3∴salt and basepresentinsoluton,soitisbasicBuffersolution.∴ salt \text{ and } basepresentinsoluton, soitisbasicBuffersolution .∴salt and basepresentinsoluton,soitisbasicBuffersolution.pOH=pkb(NH3)+logsaltbase=4.75pOH = pkb\left({NH}_{3}\right) + log \dfrac{salt}{base} = 4 .75pOH=pkb(NH3)+logbasesalt=4.75∴pOH=Pkb=4.75∴ pOH = Pkb = 4 .75∴pOH=Pkb=4.75
pH=14−pOH=14−4.75=9.25.pH = 14 - pOH = 14 - 4 . 75 = 9 . 25 .pH=14−pOH=14−4.75=9.25.
