`1)`

`n_K={3,9}/{39}=0,1(mol)`

`n_{HCl}=0,1.1=0,1(mol)`

`2K+2HCl->2KCl+H_2`

`0,1=0,1->` Phản ứng hoàn toàn

`->pH(A)=7`

`2)`

`n_{Na}={4,6}/{23}=0,2(mol)`

`n_{HCl}=0,1.1=0,1(mol)`

`2Na+2HCl->2NaCl+H_2`

`0,2>0,1->Na` dư.

`->n_{Na\ pu}=n_{HCl}=0,1(mol)`

`->n_{Na\ du}=0,2-0,1=0,1(mol)`

`2Na+2H_2O->2NaOH+H_2`

`->n_{NaOH}=n_{Na\ du}=0,1(mol)`

`->[OH^-]={0,1}/{0,1}=1M`

`->pH(A)=14+lg[OH^-]=14`

`3)`

`pH(truoc)=x`

`V_{H^+\ bd}=V(L)`

`->[H^+\ bd]=10^{-x}M`

`->n_{H^+}=10^{-x}.V(mol)`

`V_{H^+\ sau}=10V(L)`

`->[H^+\ sau]=[10^{-x}.V}/{10V}=10^{-(x+1)}M`

`->pH(sau)=-lg(10^{-(x+1)})=y`

`->x+1=y`

`4)`

`pH(truoc)=x`

`->pOH=14-x`

`->[OH^-]=[KOH]=10^{x-14}M`

`->n_{KOH}=10^{x-14}.10.10^{-3}=10^{x-16}(mol)`

`V_{sau}=10+990=1000(mL)=1(L)`

`->[OH^-]={10^{x-16}}/{1}=10^{x-16}M`

`->pOH=-lg(10^{x-16})=16-x`

`->pH(sau)=14-pOH=14-16+x=y`

`->x=y+2`