Bỏ qua sự điện li của nước.

Câu 3:

$CH_3COOH\rightleftharpoons CH_3COO^- + H^+$

Ban đầu: $C_{CH_3COOH}=0,1M$

$\Rightarrow C_{\text{phân li}}= [H^+]=[CH_3COO^-]=0,1.1,32\%=1,32.10^{-3}M$

$\Rightarrow [CH_3COOH]=0,1-1,32.10^{-3}=0,09868M$

$K_a(CH_3COOH)=\dfrac{[H^+][CH_3COO^-]}{[CH_3COOH]}=\dfrac{(1,32.10^{-3})^2}{0,09868}=1,77.10^{-5}$

Câu 4:

a,

$HClO\rightleftharpoons H^+ + ClO^-$

Gọi x là nồng độ axit phân li.

Ta có: $K_a=\dfrac{[H^+][ClO^-]}{[HClO]}$

$\Rightarrow \dfrac{x^2}{0,2-x}=4.10^{-8}$

$\Leftrightarrow x=8,94.10^{-5}$

$\Rightarrow \alpha=\dfrac{x.100}{0,2}=0,0447\%$ 

b,

Xét 1l dd axit.

$\Rightarrow m_{dd}=1000g$

$n_{HCOOH}=\dfrac{1000.0,46\%}{46}=0,1 mol$

$\Rightarrow C_{0_{\text{HCOOH}}}=0,1M$

$HCOOH\rightleftharpoons HCOO^- + H^+$

$[H^+]=10^{-3}=0,001M= C_{\text{phân li}}$

$\Rightarrow \alpha=\dfrac{0,001.100}{0,1}=1\%$