A 0.1 aqueous solution of a weak acid is 2 % ionised. If the ionic product of water is 1 × 10-4, the [ OH -]is

Question

A 0.1 aqueous solution of a weak acid is 2 % ionised. If the ionic product of water is 1 × 10-4, the [ OH -]is (A)  5× 10-12M (B)  2× 10-3M (C)  1× 10-14M  (D) None of these

Tardigrade Admin · Accepted Answer

Since, it is a weak acid, the equation to calculate [H+]is =C × α(α= % of ionisation ) =0.1 × 0.02=2 × 10-3 M Kw=[H+][O H-] [O H-]=(Kw/[H+])=(1 × 10-14/2 × 10-3) =5 × 10-12 M

Q. A $0.1$ aqueous solution of a weak acid is $2%$ ionised. If the ionic product of water is $1 \times 1 0^{- 4}$ , the $[O H^{-}]$ is

$5 \times 1 0^{- 12} M$

$2 \times 1 0^{- 3} M$

$1 \times 1 0^{- 14} M$

None of these

Since, it is a weak acid, the equation to calculate $[H^{+}]$ is $= C \times α (α = %$ of ionisation $)$
$= 0.1 \times 0.02 = 2 \times 1 0^{- 3} M$
$K_{w} = [H^{+}] [O H^{-}]$
$[O H^{-}] = \frac{K _{w}}{[ H ^{+} ]} = \frac{1 \times 1 0 ^{- 14}}{2 \times $1 0 ^{- 3}}$
$= 5 \times 1 0^{- 12} M$

Q. A 0.1 aqueous solution of a weak acid is 2% ionised. If the ionic product of water is 1×10−4, the [OH−]is

5×10−12M

2×10−3M

1×10−14M