Ai giải giúp e với ạ

$3,$

$a,$

$n_{NaOH}=n_{Ba(OH)_2}=0,01(mol)$

$n_{H_2SO_4}=0,015(mol)$

$n_{HCl}=0,005(mol)$

`->`$n_{OH^-}=0,03(mol);n_{H^+}=0,035(mol)$

$H^++OH^-  \to H_2O$

`->`$n_{H^+}=0,005(mol)$

`=>`$pH=-lg(\dfrac{0,005}{0,4+0,1})=2$

$b,$

$[H^+](gd)=\sqrt{0,1 \cdot 1,78 \cdot 10^{-4}}=10^{-2,3748}>100[OH^-]=10^{-9,6252}$

$[H^+](gd) > \dfrac{C_a}{100}=10^{-3}$

`->`$[H^+]^2+[H^+] \cdot K_a-K_aC_a=0$
`=>`$[H^+]=10^{-2,384}(M)$

`->`$pH=2,384.$

$4,$

$a,$

$[H^+](gd)=\sqrt{0,1 \cdot 10^{-4,75}}=10^{-2,875}>100[OH^-]=10^{-9,125}$

$[H^+](gd) > \dfrac{C_a}{100}=10^{-3}$

`->`$[H^+]^2+[H^+] \cdot K_a-K_aC_a=0$
`=>`$[H^+]=10^{-2,878}(M)$

`->`$pH(A)=2,878.$

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

$pH(B)=14+lg[OH^-]=14-2=12$

$b,$

$n_{H^+}=10^{-2,878} \cdot 0,1=10^{-3,878}(mol)$

$n_{OH^-}=0,01 \cdot 0,1=10^{-3}(mol)$

$OH^-+H+ \to H_2O$

`=>`$n_{OH^-,dư}=10^{-3,062}(mol)$

`=>`$[OH^-]=10^{-2,363}(M)$

`=>`$pH=14+lg(10^{-2,363})=11,637$