Question

When 900 ml of water is added in 100 ml of 0.01N $HCl$ solution then pH of resultant solution will be:

Answer 1

Given that

Volume of $HCl=100ml=0.1L$

Normality of $HCl=0.1N$

Now we can find

Gram-equivalent of $HCl=Normality\times Volume$

$=0.1\times 0.1=0.01$

So,

Volume of solution $=$Volume of $H_{2}O+$ Volume of $HCl$

$=900+100=1000mL$

$=100mL=1L$

Normality of $HCl$ in resulting solution $=\frac{0.01}{1}=0.01N$

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

$=-\log \left(0.01\right)$

$=2$

This is required solution .