logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
0 votes

Percentage ionisation of a weak monobasic acid can be calculated by using the formula

(a) $100 \sqrt{\frac{k_g}{C}}$

(b) $\frac{100}{1 + 10^{(pK_a - pH)}}$

(c) Both (a) and (b)

(d) None of these

Can you answer this question?
 
 

1 Answer

0 votes
Answer: $\frac{100}{10^{pK_a-pH}+1}$
 
$HA \rightleftharpoons H^+ + A^-$
 
Initially, [HA] = C, [$H^+$] = 0, [$A^-$] = 0
$[HA] = C(1-\alpha), [H^+] = C\alpha , [A^-] = C\alpha$
 
Ionic Product of the acid HA, $K_a = \frac{[H^+][A^-]}{[HA]} \approx C\alpha^2$ Since $1-\alpha \approx 1$
$\Rightarrow \alpha = \sqrt{\frac{K_a}{C}}$ and $C = \frac{K_a}{\alpha^2}$
 
% $\alpha = 100 \sqrt{\frac{K_a}{C}}$
 
If $\alpha$ is not very small
$K_a = \frac{C \alpha^2}{1-\alpha} = \frac{[H^+] \alpha}{1 - \alpha}$
$\Rightarrow \frac{1-\alpha}{\alpha} = \frac{[H^+]}{K_a}$
$\Rightarrow \frac{1}{\alpha} - 1 = \frac{H^+}{K_a}$
$\Rightarrow \frac{1}{\alpha} = \frac{H^+}{K_a} + 1$
$\Rightarrow \frac{1}{\alpha} = \frac{10^{-pH}}{10^{-pK_a}} + 1$
$\Rightarrow \frac{1}{\alpha} = 10^{pK_a - pH} + 1$
$\Rightarrow \frac{1}{\alpha} = \frac{1}{10^{pK_a-pH} +1}$
 
$\therefore$ % $\alpha = \frac{100}{10^{pK_a-pH}+1}$

 

answered Nov 29, 2013 by mosymeow_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...