Thank you for reporting, we will resolve it shortly
Q.
Given $pH$ of a solution $A$ is $3$ and it is mixed with another solution $B$ having $pH$ 2. If both mixed, then resultant pH of the solution will be :
$pH$ of $A=3\,\,\, pH$ of $B=2$
$\therefore \left[ H ^{+}\right]=10^{-3} M, \,\,\,\left[ H ^{+}\right]=10^{-2} M$
Total $\left[ H ^{+}\right]$
$=10^{-3}+10^{-2}=10^{-3}+10 \times 10^{-3}$
$=11 \times 10^{-3}$
$\therefore pH =-\log _{10}\left[ H ^{+}\right]=-\log \left(11 \times 10^{-3}\right)$
$=3-\log 11$
$=3-1.04=1.96$