a,
$D= 1,05g/cm^3= 1,05g/ml$
$\Rightarrow m_{CH_3COOH}=1,05.10=10,5g$
$n_{CH_3COOH}= 0,175 mol$
$\Rightarrow C_M=\frac{0,175}{1,75}=0,1M$
b,
$[H^+]= 10^{-2,9}$
$\Rightarrow [OH^-]=[H^+]_{H_2O}=\frac{K_W}{[H^+]}= 10^{-11,1}$ (rất nhỏ, bỏ qua sự phân li của nước)
$CH_3COOH\rightleftharpoons CH_3COO^- + H^+$
$\Rightarrow [CH_3COOH]_{\text{điện li}}= 10^{-2,9}$
$\alpha=\frac{10^{-2,9}.100}{0,1}= 1,26\%$
c,
$[CH_3COOH]_{\text{cân bằng}}= 0,1-10^{-2,9}=1,26.10^{-3}$
$[CH_3COO^-]=[H^+]=10^{-2,9}$
$\Rightarrow K_a= \frac{10^{-2,9}.10^{-2,9}}{1,26.10^{-3}}= 1,26.10^{-3}$
