Mình cần lời giải vs đáp án

Cần gấppppp

$10,D$

$n_{H_2}=10^{-3}(mol)$

`-` Bảo toàn $e:$

$n_{OH}=2n_{H_2}=2 \cdot 10^{-3}(mol)$

`=>`$[OH^-]=\dfrac{2 \cdot 10^{-3}}{0,2}=0,01(M)$

`=>`$pH=14+lg[OH^-]=12$

$11,D$

$n_{H^+}=n_{HCl}+n_{HNO_3}=0,01+0,01=0,02(mol)$

$V=100+100=200(mol)=0,2(l)$

$[H^+]=0,1(M)$

`=>`$pH=1$

$12,$ 

$n_{OH^-}=n_{NaOH}+n_{KOH}=0,003+0,002=0,005(mol)$

$V=300+200=500(mol)=0,5(l)$

$[OH^-]=0,01(M)$

`=>`$pH=14+lg[OH^-]=12$

$13,C$

$pH=12$`=>`$[OH^-]=10^{-2}(M)$

$n_{Ba(OH)_2}=0,0015(mol)$

`=>`$n_{OH^-}=0,2 \cdot 10^{-2}+0,0015 \cdot 2=0,005(mol)$

`=>`$[OH^-]=\dfrac{0,005}{0,5}$

$[OH^-]=0,01(M)$

`=>`$pH=14+lg[OH^-]=12$

$14,B$

$n_{HCl}=0,004(mol)$

$n_{NaOH}=0,001(mol)$

`=>`$n_{HCl,dư}=0,004-0,001=0,003(mol)$

$V=0,2+0,1=0,3(l)$

`=>`$[H^+]=0,01(M)$

`=>`$pH=2$