Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

7.59 The ionization constant of propanoic acid is $1.32\times 10^{-5}$ . Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Answers (1)

best_answer

Let the degree of ionization of propanoic acid be $\alpha$ . Then Let suppose we can write propanoic acid to be HA,

It is known that,

We have
ionization constant of propanoic acid ( $K_a$ )= $1.32\times 10^{-5}$ and the concentration is 0.005 M

$\alpha = \sqrt{\frac{K_a}{C}}$
By putting the values in above formula we get,

$\alpha = \sqrt{\frac{1.32\times 10^{-5}}{0.05}} = 1.62\times 10^{-2}$
[Acid] = $[H_{3}O]^+$ = C. $\alpha$ = $8.15\times 10^{-4}$

Therefore, $p^H = -\log [H_{3}O^+]$
$\\= -\log [8.15\times 10^{-4}]$
$=3.08$

If we add 0.01M hydrochloric acid then,

$AH+H_{2}O\rightleftharpoons H_{3}O^+ +A^-$
initial con C 0 0
at equi. C - $x$ $\approx$ C 0.01 + $x$ $x$

Now, by using the formula of $K_a = \frac{(x)(0.01)}{C-x}$
$=\frac{(x)(0.01)}{C}$

The value of $x$ is calculated as ;
$\Rightarrow 1.32\times 10^{-5}\times \frac{0.01}{0.01} = 1.32 \times 10^{-5}$ (Degree of ionisation)

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th

anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question