JEE Main 2017 · Chemistry · Ionic Equilibrium · medium

A 0.10 M solution of a weak monobasic acid HA has pH=3.0. To 100 mL of this solution, 75 mL of 0.10 M NaOH is added. The pH of the resulting solution…

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0.10\ \mathrm{M}

solution of a weak monobasic acid HA has

\mathrm{pH}=3.0

. To

100\ \mathrm{mL}

of this solution,

75\ \mathrm{mL}

0.10\ \mathrm{M}

NaOH is added. The

\mathrm{pH}

of the resulting solution is nearly (

\log3=0.48

)

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Answer: (D) $5.5$

From

\mathrm{pH}=3

[\mathrm{H^+}]=10^{-3}

K_a=\dfrac{(10^{-3})^2}{0.10-10^{-3}}\approx1.0\times10^{-5}

\mathrm{p}K_a\approx5

. Mixing: HA

=10.0

mmol, NaOH

=7.5

mmol, so

\mathrm{A^-}=7.5

and HA left

=2.5

mmol (a buffer).

\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}=5+\log3=5.48\approx5.5

\mathrm{pH}=5.0

would be the half-neutralisation point and

4.5

results from inverting the ratio.

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake