Đáp án:
`a. pH=2,1`
`b. pH=12,3`
Giải thích các bước giải:
Bài 2:
`a.`
`n_(HNO_3)=0,2.0,005=0,001` `(mol)`
`n_(HCl)=0,3.0,01=0,003` `(mol)`
$HNO_3 \to H^++NO_3^-\\0,001~~~~~0,001~~~~~~~~~~~~~~~~(mol)$
$HCl \to H^++Cl^-\\0,003~~~0,003~~~~~~~~~~~(mol)$
`=>n_(H^+)=0,001+0,003=0,004` `(mol)`
`=>[H^+]=(0,004)/(0,2+0,3)=0,008M`
`=>pH=-log[H^+]=-log(0,008)=2,1`
`b.`
`n_(KOH)=(2,8)/56=0,05` `(mol)`
`n_(Ba(OH)_2)=(4,275)/171=0,025` `(mol)`
$KOH \to K^++OH^-\\0,05~~~~~~~~~~~~~~~~~~~~0,05~~~~~(mol)$
$Ba(OH)_2 \to Ba^{2+}+2OH^-\\0,025~~~~~~~~~~~~~~~~~~~~~~~~~~~~0,05~~~~~(mol)$
`=>n_(OH^-)=0,05+0,05=0,1` `(mol)`
`=>[OH^-]=(0,1)/5=0,02M`
`=>[H^+]=(10^-14)/(0,02)=5.10^-13M`
`=>pH=-log[H^+]=-log(5.10^-13)=12,3`
