question_answer 20)

Calculate the pH of 0.08 M solution of hypochlorous acid,\[HOCl\]. The ionization constant of the acid is\[2.5\times {{10}^{-5}}\]. Determine the percentage dissociation of\[HOCl\].

Answer:

For monobasic acid, \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}=\sqrt{\frac{2.5\times {{10}^{-5}}}{0.08}}}\] \[=0.0176\] % Dissociation = 0.0176 x 100 = 1.76 \[pH=-\log C\alpha =-\log (0.08\times 0.0176)\] =2.85