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

Tardigrade Admin · Accepted Answer

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

Q. $A 0.1$ aqueous solution of a weak acid is $2$ ionized. If the ionic product of water is $1 \times 10^{- 4},$ the $[O H^{-}]$ is

$5 \times 10^{- 12} M$

$2 \times 10^{- 3} M$

$1 \times 10^{- 4} M$

None of these

Since, it is a weak acid, the equation to calculate $[H^{+}]$ is $= C \times α (α =$ of ionization) $= 0.1 \times 0.02 = 2 \times 10^{- 3} M$ $K_{w} = [H^{+}] [O H^{-}]$ $[O H^{-}] = \frac{K _{w}}{[ H ^{+} ]} = \frac{1 \times 10 ^{- 14}}{2 \times 10 ^{- 3}}$ $= 5 \times 10^{- 12} M$

Q. A0.1 aqueous solution of a weak acid is 2 ionized. If the ionic product of water is 1×10−4, the [OH−] is

5×10−12M

2×10−3M

1×10−4M