Question

$100ml$ of a solution of $HCl$ with $pH$ value $3$ is diluted with $400ml$ of water. The new $pH$ of the solution is.

• A. $3.7$
• B. $5.3$
• C. $4.2$
• D. $5.6$

Answer 1

The correct option is A $3.7$

$pH=-log\left[H^{+}\right]=3$

So, $\left[H^{+}\right]=0.001M$

$V_1\times N_1=V_2\times N_2$

$\left(100ml\right)\times \left(0.001M\right)=\left(500ml\right)\times N_2$

$N_2=0.0002M$

$pH=-log\left[H^{+}\right]=log\left(0.0002\right)=3.7$.

Option A is correct.