An aqueous solution of a weak monoprotic acid, HA , has a concentration of 0.02 ~M . If the acid is 0.05 % dissociated at 298 ~K , the pH of the solution is :

Step-by-step solution

Given concentration of the weak monoprotic acid, C = 0.02 ~M Degree of dissociation, = 0.05 % = 0.05 100 = 5 10⁻⁴ For a weak monoprotic acid, the concentration of hydrogen ions is given by: [H^+] = C Substituting the given values: [H^+] = 0.02 5 10⁻⁴ = 10⁻⁵ ~M The pH of the solution is calculated as: pH = - [H^+] pH = - (10⁻⁵) = 5 Answer: 5