Tính pH của dd CH3COOH 0,3M, biết Ka = 1,8.10^-5 ? Nếu cho thêm 0,84 g KOH vào 500ml dd trên. Tính pH của dd thu đc

`@` Phương trình ion :

$\hspace{1,5cm}\rm CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \hspace{1,5cm} (K_a=1,8.10^{-5})$$\\$Ban đầu:$\rm \hspace{1cm}0,3\hspace{2.3cm}0\hspace{1,6cm}0\hspace{2cm}(M)$$\\$Phân li: $\rm \hspace{1,5cm}x\hspace{0,5cm}\to\hspace{1,2cm}x\hspace{0.5cm}\to\hspace{0.3cm}x\hspace{2cm}(M)$$\\$Cân bằng:$\rm \hspace{0,2cm}(0,3-x)\hspace{1,8cm}x\hspace{1,5cm}x\hspace{1,9cm}(M)$

`K_a=([CH_3COO^-][H^+])/([CH_3COOH])=(x.x)/(0,3-x)=1,8.10^{-5}`

`=>x≈2,315.10^{-3}(M)`

`pH=-log[H^+]=-log[2,315.10^{-3}]≈2,635`

`@` `n_{KOH}=(0,84)/(56)=0,015(mol)`

`n_{CH_3COOH}=0,5.0,3=0,15(mol)`

Phản ứng :

`CH_3COOH+KOH\to CH_3COOK+H_2O`

`0,015\leftarrow 0,015(mol)`

`=>n_{CH_3COOH(\text{dư})}=0,15-0,015=0,135(mol)`

`=>` Nồng độ mol : $\begin{cases} CH_3COOK:0,015:0,5=0,03(M)\\CH_3COOH:0,135:0,5=0,27(M) \end{cases}$

Phương trình phân li :

$\rm CH_3COOK\to CH_3COO^- +K^+$

$\rm \,\,\,0,03 \to 0,03(M)$

$\hspace{1,5cm}\rm CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \hspace{1,5cm} $$\\$Ban đầu:$\rm \hspace{1cm}0,27\hspace{2.2cm}0\hspace{1,5cm}0\hspace{2cm}(M)$$\\$Phân li: $\rm \hspace{1,5cm}x\hspace{0,5cm}\to\hspace{1,2cm}x\hspace{0.5cm}\to\hspace{0.3cm}x\hspace{2cm}(M)$$\\$Cân bằng:$\rm \hspace{0,1cm}(0,27-x)\hspace{1,8cm}x\hspace{1,5cm}x\hspace{1,9cm}(M)$

`K_a=([CH_3COO^-][H^+])/([CH_3COOH])=((x+0,03).x)/(0,27-x)=1,8.10^{-5}`

`=>x≈1,61.10^{-4}`

`=>pH=-log[H^+]=-log[1,61.10^{-4}]≈3,793`

Phương trình điện li 

$\rm ~~~~~~~~~~~~~~~CH_3COOH \rightleftharpoons H^+ +CH_3COO^-\\ ban~đầu~~~~~0,3~~~~~~~~~~~~~~~~~0~~~~~~~~~~~~~0\\phản~ứng ~~~~~a~~~~~~~~~~~~~~~~~~~a~~~~~~~~~~~~~a\\~~[]~~~~~~~~~~(0,3-a)~~~~~~~~~~~~~a~~~~~~~~~~~~~a$

Ta có $\rm K_a=\dfrac{[H^+][CH_3COO^-]}{[CH_3COOH]}$ 

`<=>`$\rm 1,8.10^{-5}=\dfrac{a^2}{0,3-a}$ 

`<=>`$\rm x=2,33.10^{-3} M$ 

`->`$\rm pH=-log[H^+]=-log(2,33.10^{-3})=2,633$ 

$\rm n_{KOH}=\dfrac{0,84}{56}=0,015(mol)$ 

$\rm n_{CH_3COOH}=0,3.0,5=0,15(mol)$ 

Phương trình ion 

$CH_3COOH  + OH^- \to CH_3COO^- +H_2O$ 

Ta có $\rm n_{CH_3COOH(dư)}=n_{CH_3COOH}-n_{OH^-}=0,15-0,015=0,135(mol)$ 

`->`$\rm [CH_3COOH(dư)]=\dfrac{0,135}{0,5}=0,27M$ 

$\rm [CH_3COO^-]=\dfrac{0,015}{0,5}=0,03M$ 

Áp dụng phương trình Henderson-Hasselbalch

$\rm pH=pK_a + log( \dfrac{[CH_3COO^-]}{[CH_3COOH]})$ 

`<=>`$\rm pH=-log(1,8.10^{-5})+ \log(\dfrac{0,03}{0,27})$ 

`<=>`$\rm pH=3,79$