JEE Advanced 2022 Paper 2 · Chemistry · Ionic Equilibrium · medium

Equal volumes (50.0 mL each) of 0.20 M weak acid HA (Ka=1.0×10^-5) and 0.20 M weak base BOH (Kb=1.0×10^-3) are mixed at 298 K (Kw=1.0×10^-14). Treat…

Numerical answer

Equal volumes (

50.0\ \mathrm{mL}

each) of

0.20\ \mathrm{M}

weak acid

\mathrm{HA}

(

K_a=1.0\times10^{-5}

) and

0.20\ \mathrm{M}

weak base

\mathrm{BOH}

(

K_b=1.0\times10^{-3}

) are mixed at

298\ \mathrm{K}

(

K_w=1.0\times10^{-14}

). Treat the resulting

0.10\ \mathrm{M}

solution of the salt

\mathrm{BA}

as a dilute ideal solution. The pH of the final solution, rounded to the nearest integer, is _____.

Show answer and solution

Answer: 8

Equal moles of weak acid and weak base neutralise completely and leave only the salt

\mathrm{BA}

0.10

M. Both ions hydrolyse:

\mathrm{A^-}+\mathrm{H_2O}\rightleftharpoons\mathrm{HA}+\mathrm{OH^-}

and

\mathrm{B^+}+\mathrm{H_2O}\rightleftharpoons\mathrm{BOH}+\mathrm{H^+}

.

Writing the proton balance

[\mathrm{H^+}]+[\mathrm{HA}]=[\mathrm{OH^-}]+[\mathrm{BOH}]

and using

[\mathrm{HA}]=[\mathrm{A^-}][\mathrm{H^+}]/K_a

and

[\mathrm{BOH}]=[\mathrm{B^+}]K_w/(K_b[\mathrm{H^+}])

, with both ion concentrations

\approx c

and hydrolysis small, gives the standard result independent of concentration:

[\mathrm{H^+}]=\sqrt{\dfrac{K_wK_a}{K_b}}=\sqrt{\dfrac{10^{-14}\times10^{-5}}{10^{-3}}}=10^{-8}

.

Hence

\mathrm{pH}=7+\tfrac12(\mathrm{p}K_a-\mathrm{p}K_b)=7+\tfrac12(5-3)=8

. A numerical solution of the full proton balance gives

7.998

, confirming the approximation. The answer is

8

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