4.1
$n_{CH_3COOH}=0,15.0,1=0,015(mol)$
$n_{NaOH}=0,1.0,1=0,01(mol)$
$CH_3COOH+NaOH\to CH_3COONa+H_2O$
Sau phản ứng:
$V_{dd}=0,15+0,1=0,25l$
$n_{CH_3COO^-}=0,01(mol)\Rightarrow C^0_{CH_3COO^-}=0,04M$
$n_{CH_3COOH}=0,015-0,01=0,005(mol)\Rightarrow C^0_{CH_3COO^-}=0,02M$
$CH_3COOH\rightleftharpoons CH_3COO^-+H^+$
Đặt $[H^+]=x(M)$
$[CH_3COO^-]=0,02+x(M)$
$[CH_3COOH]=0,04-x (M)$
$\Rightarrow \dfrac{x(0,02+x)}{0,04-x}=1,75.10^{-5}$
$\Leftrightarrow x^2+0,02x=1,75.10^{-5}(0,04-x)$
$\Leftrightarrow x=3,5.10^{-5}$
$\to pH=-\log_{10}x=4,46$
