`n_(OH^-)=n_(NaOH)=0,2.0,002=4.10^-4 (mol)`
`n_(H^+)=n_(HCl)=0,4.0,01=4.10^-3 (mol)`
`NaOH+HCl->NaCl+H_2O`
Lập tỉ lệ
`n_(OH^-)=\frac{4.10^4}{1}<n_(H^+)=\frac{4.10^-3}{1}`
`=>H^+` dư
`[H^+]=\frac{3,6.10^-3}{0,6}=6.10^-3`
`pH=-log(6.10^-3)=2,22`
`b,`
`n_(H^+)=\frac{3,6.10^-3}{5}+0,2.1=0,20072(mol)`
`[H^+]=\frac{0,20072}{0,32}=0,62725M`
`pH=-log(0,62725)=0,2`
`H^(+)+OH^(-)->H_2O`
`0,20072` `0,20072`
`=>V_(KOH)=\frac{0,20072}{0,5}=0,40144(l)=401,44(ml)`
`d,`
Dung dịch `Y` có dư `H^+` nên `H_2SO_4` không thể trung hòa :))