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Q.
$4 \,ml$ of $HCl$ solution of $pH =2$ is mixed with $6\, ml$ of $NaOH$ solution of $pH =12$. What would be the final $pH$ of solution $? \log 2=0.3$
Equilibrium
Solution:
$pH =2,[ HCl ]=10^{-2}( M ) \& pOH =2$
or $[ NaOH ]=10^{-2}( M )$
$4\, ml$ of $10^{-2}$ (M) $HCl \equiv 4 \times 10^{-5}$ moles $HCl$.
$6 \,ml$ of $10^{-2}$ (M) $NaOH \equiv 6 \times 10^{-5}$ moles $NaOH$
After mixing excess moles of $OH ^{-}=2 \times 10^{-5}$
$\left[ OH ^{-}\right]=\frac{2 \times 10^{-5}}{10} \times 10^{3}=2 \times 10^{-3}$
or $pOH =3-\log 2=3-0.3=2.7$
or $pH =11.3$