JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Equilibrium

50 mL of

0.1 M CH_{3} COOH

is being titrated against 0.1 M NaOH . When 25 mL of NaOH has been added, the pH of the solution will be
____

\times 1 0^{- 2}

. (Nearest integer)
(Given :

pK_{a} (CH_{3} COOH) = 4.76

)

lo g 2 = 0.30

lo g 3 = 0.48

lo g 5 = 0.69

lo g 7 = 0.84

lo g 11 = 1.04

Answer: 476

Solution

Initial moles of

CH X_{3} COOH

50 \times 0.1 = 5

mmol

Moles of

NaOH

added =

25 \times 0.1 = 2.5

mmol

After neutralization:

This is a buffer solution at the half-equivalence point. Using Henderson-Hasselbalch equation:

pH = pK_{a} + lo g (\frac{[ CH X _{3} COO X ^{-} ]}{[ CH X _{3} COOH ]})

pH = 4.76 + lo g (\frac{2.5}{2.5}) = 4.76 + 0 = 4.76

Expressed as

\times 1 0^{- 2}

4.76 = 476 \times 1 0^{- 2}

Answer: 476

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer