JEE Main 2019 · Chemistry · Ionic Equilibrium · medium

100 mL of 0.10 M acetic acid is mixed with 40 mL of 0.10 M NaOH. If pKa of acetic acid is 4.76, the pH of the resulting solution is nearest to (log 2…

Single correct option

100\ \mathrm{mL}

0.10\ \mathrm{M}

acetic acid is mixed with

40\ \mathrm{mL}

0.10\ \mathrm{M}

NaOH. If

\mathrm{p}K_a

of acetic acid is

4.76

, the pH of the resulting solution is nearest to (

\log 2 = 0.30

\log 3 = 0.48

)

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Answer: (A) $4.58$

Moles: acid

=10

mmol, NaOH

=4

mmol. NaOH converts 4 mmol of acid into 4 mmol of acetate, leaving

6

mmol acid. This is a buffer:

\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} = 4.76 + \log\dfrac46 = 4.76 - (0.48-0.30) = 4.58

. Volumes cancel because both species share the same final volume. Inverting the ratio gives the wrong value

4.94

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake