Đáp án:

$\rm a)\\n_{H^+}=10^{-3}\ (M)$

Gọi a là nồng độ mol của dd acid acetic.

Phương trình điện ly:

$\rm CH_3COOH\leftrightarrows CH_3COO^-+H^+\\\kern,3cma-10^{-3}\kern1,8cm 10^{-3}\kern1cm 10^{-3}\kern,5cm (M)$

$\rm K_a=\dfrac{ [CH_3COO^-]×[H^+]}{ [CH_3COOH]}=\dfrac{(10^{-3})^2}{a-10^{-3}}=1,75×10^{-5}$

$\rm\to C_{M\ CH_3COOH}=\dfrac{407}{7000}\ (M)$

$\rm b)$

$\rm n_{NaOH}=0,2×0,058=0,0116\ (mol)\\n_{CH_3COOH}=0,2×\dfrac{407}{7000}=\dfrac{407}{35000}\ (mol)$

Phản ứng xảy ra:

$\rm CH_3COOH+NaOH\to CH_3COONa+H_2O$

Nhận thấy: $\rm n_{CH_3COOH}>n_{NaOH}\to CH_3COOH$ dư sau phản ứng.

Theo phương trình:

$\rm n_{CH_3COONa}=n_{NaOH}=0,0116\ (mol)$

$\rm n_{CH_3COOH\ (du)}=\dfrac{407}{35000}-0,0116=2,857×10^{-5}\ (mol)$

Dung sau phản ứng gồm:

$\rm [base]=[CH_3COONa]=\dfrac{0,0116}{0,2+0,2}=0,029\ (M)$

$\rm [acid]=[CH_3COOH]=\dfrac{2,857×10^{-5}}{0,2+0,2}=7,1425×10^{-5}\ (M)$

$\rm pH=pK_a-\log\bigg( \dfrac{ [acid]}{ [base]}\bigg)\\\quad\;=-log(1,75×10^{-5})-\log\bigg(\dfrac{7,1425×10^{-5}}{0,029}\bigg)=7,3655$