Test Index

NCERT Class XI Chemistry Equilibrium Solutions

© examsnet.com
Question : 47
Total: 73
It has been found that the pH of 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Solution:  
pH = - log [H+]
∴ [H+] = antilog (- 4.15) = 7.08 × 10−5M
HA(aq) ⇌ H(aq)++A(aq)−
∴ [H+] = [A−] = 7.08 × 10−5M
and α =
[H+]
X
 =
7.08×10−5
0.01
 = 7.08 × 10−3
Ka =
Cα2
1−α
 =
0.01(7.08×10−3)2
1−7.08×10−3
 = 5.05 × 10−7
pKa = - log Ka = - log (5.05 × 10−7) = 6.29
© examsnet.com
Go to Question:
Prev Question
Next Question