1.0 M acetic acid is mixed with 0.01 mol L-1 HCl. The pH of the resultant solution is [Ka(CH3COOH) = 1.8 × 10-5]

Question

1.0 M acetic acid is mixed with 0.01 mol L-1 HCl. The pH of the resultant solution is [Ka(CH3COOH) = 1.8 × 10-5] (A) 1.846 (B) 1.04 (C) 2.36 (D) 2.79

Tardigrade Admin · Accepted Answer

For weak acid, CH3COOH, [H+]=√KaC √1.8×10-5×1 [H+]=4.242×10-3 Total [H+]=0.01 + 4.242 × 10-3 = 1.424 × 10-2 pH=-log [H+] pH = -log(1.424 × 10-2) ∴ pH=1.846

Q. $1.0 M$ acetic acid is mixed with $0.01 m o l L^{- 1} H Cl$ . The $p H$ of the resultant solution is
$[K_{a} (C H_{3} COO H) = 1.8 \times 1 0^{- 5}]$

$1.846$

$1.04$

$2.36$

$2.79$

For weak acid, $C H_{3} COO H, [H^{+}] = K_{a} C$
$1.8 \times 1 0^{- 5} \times 1$
$[H^{+}] = 4.242 \times 1 0^{- 3}$
Total $[H^{+}] = 0.01 + 4.242 \times 1 0^{- 3} = 1.424 \times 1 0^{- 2}$
$p H = - l o g [H^{+}]$
$p H = - l o g (1.424 \times 1 0^{- 2})$
$∴ p H = 1.846$

Q. 1.0M acetic acid is mixed with 0.01molL−1HCl. The pH of the resultant solution is [Ka​(CH3​COOH)=1.8×10−5]

1.846

1.04

2.36

2.79