Đáp án:

$\rm pH=3\to C_{H^+}=10^{-3}\ (M)\\pH=11\to C_{OH^-}=10^{-3}\ (M)$

Gọi $\rm C_{A}$ là nồng độ ban đầu của $\rm CH_3COOH$.

Phương trình phân li:

$\rm CH_3COOH\leftrightarrows CH_3COO^-+H^+$

Tại thời điểm cân bằng: $\begin{cases}\rm CH_3COO^-:10^{-3}\ (M)\\\rm H^+:10^{-3}\ (M)\\\rm CH_3COOH:C_A-10^{-3}\ (M)\end{cases}$

$\rm K_a=\dfrac{(10^{-3})^2}{C_A-10^{-3}}=10^{-4,76}$

$\rm\to C_A≈0,0585\ (M)$

$\rm n_{CH_3COOH}=0,0585×25×10^{-3}=1,4625×10^{-3}\ (mol)\\n_{KOH}=10^{-3}×15×10^{-3}=1,5×10^{-5}\ (mol)$

Phản ứng xảy ra:

$\rm CH_3COOH+KOH\to CH_3COOK+H_2O$

Dung dịch đệm gồm: $\begin{cases}\rm CH_3COOH:1,4475×10^{-3}\ (mol)\\\rm CH_3COOK:1,5×10^{-5}\ (mol)\end{cases}$

$\rm C_{CH_3COOH}=\dfrac{1,4475×10^{-3}}{(25+15)×10^{-3}}=0,0362\ (M)\\C_{CH_3COOK}=\dfrac{1,5×10^{-5}}{(25+15)×10^{-3}}=3,75×10^{-4}\ (M)$

$\rm pH=pK_a+\log\bigg(\dfrac{C_{CH_3COOK}}{C_{CH_3COOH}}\bigg)=4,76+\bigg(\dfrac{3,75×10^{-4}}{0,0362}\bigg)=2,775$