Question

PH = 10(900ml)
pH = 11 [100ml]
Find net pH

Answer 1

Volume of solution 1 $V_1=900mL$

$pH=10$

$p{H}_1=-\log [H^{+}{]}_1$ $p{H}_2=-\log [H^{+}{]}_2$

$10=-\log \frac{n_1}{900}$ $11=-\log \frac{n_2}{100}$

Resulting solutions,

Total volume=$1000mL$

Total no. of $\left[H^{+}\right]$ ions=$n_1+n_2$

$\left[H^{+}\right]$=$\frac{N_1+n_2}{1000}=\frac{{10}^{-10}+{10}^{-11}}{1000}$

$pH=-\log \left[\frac{{10}^{-10}+{10}^{-11}}{1000}\right]$

$=9.9586-3$

$=6.9586$