Đáp án:

`1. \ pH=13,875`

`2. \ pH=0,377`

Giải:

1.

`pH=13 → [H^+]=10^{-13}`

Nồng độ `OH^-` sau khi pha loãng:

`[OH^-]=\frac{10^{-14}}{[H^+]}=0,1 \ (M)`

→ `n_{OH^-}=[OH^-]V_2=0,1.1,5=0,15 \ (mol)`

Nồng độ `OH^-` ban đầu:

`[OH^-]=\frac{0,15}{0,2}=0,75 \ (M)`

→ `[H^+]=\frac{10^{-14}}{0,75}=1,(3).10^{-14} \ (M)`

→ `pH=-log[H^+]=-log[1,(3).10^{-14}]=13,875`

2.

`n_{HCl}=0,3.0,5=0,15 \ (mol)`

`n_{H_2SO_4}=0,2.0,15=0,03 \ (mol)`

Phương trình điện li:

`HCl → H^++Cl^-`

`H_2SO_4 → 2H^++SO_4^{2-}`

`n_{H^+}=0,15+0,03.2=0,21 \ (mol)`

`[H^+]=\frac{n_{H^+}}{V}=\frac{0,21}{0,5}=0,42 \ (M)`

→ `pH=-log[H^+]=-log0,42=0,377`

1.

$pOH=14-13=1$

$\to [OH^-]=10^{-1}=0,1M$

$V_{dd}=0,2+1,3=1,5l$

$\to n_{OH^-}=1,5.0,1=0,15(mol)$

$\to pH=14+\log[OH^-]=14+\log\dfrac{0,15}{0,2}=13,875$

2.

$V_{dd}=0,3+0,2=0,5l$

$\sum n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,21(mol)$

$\to pH=-\log\dfrac{0,21}{0,5}=0,377$