Question

Calculate the pH of a solution obtained by mixing $100mL$ of an acid of $pH=3$ and $400mL$ solution of $pH=1$.

Answer 1

$pH=3$ means $\left[H^{+}\right]={10}^{-3}$ and $pH=1$ means $\left[H^{+}\right]=0.1M$

So $M=\frac{M_1\times V_1+M_2\times V_2}{V}$

$M=\frac{{10}^{-3}\times 100+0.1\times 400}{500}=0.08$

$\left[H+\right]=0.08$

So $pH=-log\left(\right[H^{+}\left]\right)=1.096$