The electrolytic reduction of 300 mL of 0.01 M nitroalkane was carried out in an acidic buffer medium of pH 5.0, following the change: RN O 2 + 4 H 3 O + + 4 e − ⟶ RN H O H + 5 H 2 O If the tota

Question

The electrolytic reduction of 300 mL of 0.01 M nitroalkane was carried out in an acidic buffer medium of pH 5.0, following the change:

$RN O_{2} + 4 H_{3} O^{+} + 4 e^{-} ⟶ RN H O H + 5 H_{2} O$

If the total concentration of the weak acid and its conjugate base was 0.056 M, calculate the pH of the solution after completion of the reduction. The $K_{a}$ for the weak acid is $1.8 \times 1 0^{- 5}$ ( $p K_{a} = 4.74$ ).

[Given: $lo g 2 = 0.30$ , $lo g 3 = 0.48$ ]

TestHub · Accepted Answer

Milli equivalent of R NO 2 ​ = 300 × 0.01 × 4 = 12 ∴ Milli equivalent of [ H + ] consumed = 12 or Milli equivalent of [ OH − ] generated = 12 Let a mole of weak acid and b mole of its conjugate base are present, then a + b = 0.50 Also, p H = − lo g K a ​ + lo g [ Acid ] [ Salt ] ​ 5.0 = + 4.74 + lo g a b ​ ∴ a b ​ = 1.8 ∴ a = 0.020 b = 0.036 ​ OH − generated will increase the concentration of A − ion

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