Given pH of a solution ' A ' is ' 3 ' and it is mixed with another...

Given that,

$pH$ of solution $A = 3$

$\left[H^{+}\right]$ of $A = {10}^{- 3}M$ and, $pH$ of solution $B = 2$

$\left[H^{+}\right]$ of $B = {10}^{- 2}M$

Thus, total $\left[H^{+}\right]$ are $= {10}^{- 3} + {10}^{- 2} = 1. 1 \times {10}^{- 3}M$

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

$= - \log \left[1. 1 \times {10}^{- 3}\right]$

$= 3 - \log 1. 1$

$= 3 - 1. 04 = 1. 9$

Therefore, the resultant $pH$ of the solution will be 1. 9$</p>$