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 _____________ $\times$ 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 (integer)
476
Solution
<p>CH<sub>3</sub>COOH + NaOH $\to$ CH<sub>3</sub>COONa + H<sub>2</sub>O</p>
<p>After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml</p>
<p>Initially,</p>
<p>Number of millimole of NaOH = 25 $\times$ 0.1 = 2.5 mm</p>
<p>Number of millimole of CH<sub>3</sub>COOH = 50 $\times$ 0.1 = 5 mm</p>
<p>After nutrilisation,</p>
<p>Millimole of NaOH = 0</p>
<p>Millimole of CH<sub>3</sub>COOH = 5 $-$ 2.5 = 2.5 mm</p>
<p>Millimole of CH<sub>3</sub>COONa = 2.5</p>
<p>After nutrilisation,</p>
<p>Concentration of CH<sub>3</sub>COOH = $[C{H_3}COOH] = {{5 - 2.5} \over {75}} = {1 \over {30}}$</p>
<p>Concentration of CH<sub>3</sub>COONa = $[C{H_3}COONa] = {{ 2.5} \over {75}} = {1 \over {30}}$</p>
<p>${P^H} = {P^{Ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$</p>
<p>$= 4.76 + \log {{{1 \over {30}}} \over {{1 \over {30}}}}$</p>
<p>$= 4.76 + \log (1)$</p>
<p>$= 4.76 + 0$</p>
<p>$= 4.76$</p>
<p>$= 4.76 \times {10^{ - 2}}$</p>
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
This question is part of PrepWiser's free JEE Main question bank. 67 more solved questions on Equilibrium are available — start with the harder ones if your accuracy is >70%.