1)

n_{NaOH}={200.1,08.0,37\%}/{40}≈0,02(mol)

->n_{OH^-}=n_{NaOH}=0,02(mol)

->[OH^-]={0,02}/{0,2}=0,1M

->pH=-log({10^{-14}}/{[OH^-]})=13

2)

pH(sau)=13

->[OH^-\ sau]={10^{-14}}/{10^{-13}}=0,1M

->n_{OH^-\ bd}=0,1.(0,2+1,3)=0,15(mol)

->[OH^-\ bd]={0,15}/{0,2}=0,75M

->pH(bd)=-log({10^{-14}}/{[OH^-]})≈13,88

3)

Đặt V_{dd\ HCl}=x(L)

n_{H_2SO_4}=0,1.0,05=0,005(mol)

pH(HCl)=2->[H^+(HCl)]=0,01M

->\sum n_{H^+}=0,01x+0,005.2=0,01x+0,01(mol)

pH(sau)=1,2->[H^+\ sau]=10^{-1,2}M

->\sum n_{H^+}=10^{-1,2}.(x+0,1)=0,01x+0,01

->x≈0,0695(L)=69,5(mL)

-> Cần 69,5mL\ HCl