Question

50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be ____ × 10–2. (Nearest integer)
(Given : pKa (CH3COOH) = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04

Answer 1

$CH_3COOH + NaOH → CH_3COONa + H_2O$

at initially 50×0.1mmoles 25×0.1mmoles
at time t 2.5 m moles 0 2.5 m mol

$pH=pK_a+log|\frac {(salt)}{(acid)}|$

$pH=4.76+log|\frac {2.5}{2.5}|$
$pH = 4.76$
$pH = 476 \times 10^{-2}$
Given that, the pH of the solution will be $x × 10^{–2}.$
Then, $x = 476$

So, the answer is $476$.