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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 53
Total: 73
Calculate the degree of ionization of 0.05 M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains
(a) 0.01 M
(b) 0.1 M HCl?
Solution:  
pKa = 4.74
–log Ka = 4.74
Ka = antilog(–4.74) = 1.820 × 10–5
Cα2 = 1.82 × 10–5
⇒ α2 =
1.82×10−5
C
 =
1.82×10−5
0.05
 = 3.64 × 10−4 ⇒ α = 1.9 × 10−2
(a) CH3COOH ⇌ CH3COO−+H+
Initial‌molar‌conc.C00
Eqm.molar‌conc.C(1−α)CαCα

Ka =
[CH3COO−][H+]
C
[H+] = 0.01 M
1.82 × 10−5 =
Cα
C
 × 0.01 ⇒ α =
1.82×10−5
0.01
 = 1.82 × 10−3
(b) [H+] = 0.1 M
1.82 × 10−5 =
Cα
C
 × 0.1 = α × 0.1 ⇒ α = 1.82 × 10−4
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