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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 46
Total: 73
The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.
Solution:  
α = √
Ka
C
 = √
1.74×10−5
5×10−2
 = 0.0186
CH3COOH ⇌ CH3COO−+H+
Ka =
[CH3COO−][H+]
[CH3COOH]
 =
[H+]2
[CH3COOH]
or, [H+] = √Ka[CH3COOH] = √(1.74×10−5)(5×10−2) = 9.33 × 10−4M
[CH3COO−] = [H+] = 9.33 × 10−4M
pH = - log (9.33 × 10−4) = 4 - 0.9699 = 4 - 0.97 = 3.03
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