Đáp án:
$\begin{array}{l} 1. & pH=1,632 \\ 2. & pH=12,3 \end{array}$
Giải:
1.
`n_{HCl}=0,01.0,1=0,001 \ (mol)`
`n_{HNO_3}=0,03.0,2=0,006 \ (mol)`
`n_{H^+}=n_{HCl}+n_{HNO_3}=0,001+0,006=0,007 \ (mol)`
`[H^+]=\frac{0,007}{0,3}=0,02(3) \ (M)`
`pH=-log[H^+]=1,632`
2.
`n_{NaOH}=0,01.0,1=0,001 \ (mol)`
`n_{KOH}=0,03.0,1=0,003 \ (mol)`
`n_{OH^-}=n_{NaOH}+n_{KOH}=0,001+0,003=0,004 \ (mol)`
`[OH^-]=\frac{0,004}{0,2}=0,02 \ (M)`
`pOH=-log[OH^-]=1,7`
`pH=14-pOH=12,3`