Question

If a solution of $pH=5$ is diluted $100$ times, its $pH$ becomes:

• A. $7.3010$
• B. $7$
• C. $6.699$
• D. $6.3010$

Answer 1

The correct option is C $6.699$

A strong acid of $pH=5$ can be considered as $1$ litre of pure water, in which ${10}^{-5}$ moles of a strong acid have been dissolved.

The $pH=5$ solution diluted $100$ times is actually a ${10}^{-7}$M solution.

But this time, we are considering the dissociation of acid only.

Now, when water is added, water also dissociates into $H^{+}$ and $O{H}^{-}$.

So, we have to account for the total concentration of $H^{+}$ from acid (${10}^{-7}M$) and from water(${10}^{-7}M$).

In case of infinite dilution of an acid, it is $pH=7$.

$\left[H^{+}\right]={10}^{-7}+{10}^{-7}=2\times {10}^{-7}$

$pH=-log(2\times {10}^{-7})=7-0.301=6.699$

Option C is correct.