Question

The pH of a $0.002N$ acetic acid solution if it is $2.3$% ionised and this dilution is : $(\log 4.6=0.6628)$

• A. $4.3372$
• B. $0.4337$
• C. $3.4337$
• D. $0.6628$

Answer 1

The correct option is A $4.3372$

Concentration $=C=0.002N$

$C=0.02N=(1\times 0.002)M=0.002M$

Dissociation is $\alpha =0.023$

After ionising, the soluiton will be acidic and contain $H^{+}$ ions.

$\left[H^{+}\right]=C\alpha =0.002\times 0.023=4.6\times {10}^{-5}$

From the $pH$ fomrula, we find $pH$

$\Rightarrow pH=-log\left[H^{+}\right]=-log(4.6\times {10}^{-5})=5-log\left(4.6\right)=5-0.6628=4.3372$

$\Rightarrow pH=4.3372$