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.
$\Rightarrow m_{dd}=1000g$
$n_{HCOOH}=\dfrac{1000.0,46\%}{46}=0,1 mol$
$\Rightarrow C_{0_{\text{HCOOH}}}=0,1M$
$HCOOH\rightleftharpoons HCOO^- + H^+$
$[H^+]=10^{-3}=0,001M= C_{\text{phân li}}$
$\Rightarrow \alpha=\dfrac{0,001.100}{0,1}=1\%$
