where α = Degree of dissociation
Dissociation constant of acid,
Ka = \(\frac{CH_3COO^-][H^+]}{[CH_3COOH]}\)
= \(\frac{Ca.Ca}{C(1-a)}\)
= \(\frac{C^2a^2}{C(1-a)}\)
∵ For weak acid (1-a) is negligible. So,
Ka = \(\frac{Ca^2}{1}\)
Ka = Ca2
a = \(\sqrt{\frac{K_a}{C}}\)
= \(\sqrt{{1.8\times10^{-5}}\over0.01}\)
= \(\sqrt{18\times10^{-4}}\)
= 4.24 x 10-2
[H+] = Ca
= 0.01 x 4.24 x 10-2
= 4.24 x 10-4
pH = −log10[H+]
= −log10[4.24 × 10−4 ]
= −log104.24 + 4 log10
= −0.6273 + 4 = 3.37
