Question

Arrange the 0.15 M solutions of the following compounds in order of increasing acidity (least acidic first).

NH₄I                B. CH₃COONH₄          C. CH₃COOK           D. KCl                 E. CH₃COOH          

         F. KF                   G. HBr                          H. CHCl₂COOH        I. NH₃                  J. CF₃COOH

      ammonia,    K_b = 1.80

Answer 1

Lower the value of pK_a stronger is the acid and similarly lower the value of pK_b stronger is the base. The relation between K_a and pK_a and K_b and pK_b is given by the following expression,

From question, it is given that

Substituting these values in equations (i) and (ii), the pK_a and pK_b values can be calculated as follows:

Since, the pK_b value of NH₃ is higher therefore it is strong base and hence a weak acid, CH₃COOK is a salt of weak acid (CH₃COOH) and strong base (KOH) so, it is a basic and hence itis a weak acid but stronger than ammonia, KF is an alkali halide so, it is basic and hence weak acid but stronger than CH₃COOK, CH₃COONH₄ is formed by a weak acid (CH₃COOH) and a weak base (NH₄OH) and KCl is formed by a strong base and strong acid so both CH₃COONH₄ and KCl have equal acidity but stronger than KF, NH₄I is formed by a strong acid (HI) and a moderate base therefore, its acidity is greater than CH₃COONH₄ = KCl, CH₃COOH is a weak acid but stronger than NH₄I, as the number of halogen atoms increases by replacing the hydrogen atoms in the acetic acid the acidity of the acid increases, therefore, CHCl₂COOH is more acidic than CH₃COOH and CF₃COOH is more acidic than CHCl₂COOH. Since, HBr is highly acidic hence it has the highest acidity among the given compounds. Hence, the correct order of acidity in increasing order is

NH₃ < CH₃COOK < KF < CH₃COONH₄ = KCl < NH₄I < CH₃COOH < CHCl₂COOH < CF₃COOH < HBr