22/
$n_{OH^-}= 2n_{Ca(OH)_2}=2.0,25.1=0,5 mol$
$n_{H^+}=n_{HNO_3}+n_{HCl}=0,35.1+0.35.2=1,05 mol$
$\Rightarrow n_{H^+\text{dư}}=1,05-0,5=0,55 mol$
$\Rightarrow [H^+]=\dfrac{0,55}{0,25+0,35}=\dfrac{11}{12}M$
$pH=-\lg \dfrac{11}{12}=0,04$
23/
$pOH=14-13=1$
$\Rightarrow [OH^-]=10^{-1}=0,1M$
$C_{M_{Ba(OH)_2}}= 0,5.0,1=0,05M$
$\Rightarrow n_{Ba}=n_{Ba(OH)_2}=0,05.2=0,1 mol$
$m=0,1.137=13,7g$