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