Calculation of Hydrolysis Constant, Degree of Hydrolysis and pH of Salt Solution

Salt of weak Acid and strong base

a) Hydrolysis Constant

 

b) Degree of Hydrolysis

A‾ + H₂O OH‾ + HA

c                       0         0             original molar conc

c(1-h )             ch         ch          Molar conc at equilibrium

c) pH

 

Salt of strong Acid and weak base

a) Hydrolysis Constant

BA + H₂O  BOH + HA

B⁺ + A‾ + H₂O  BOH + H⁺ + A‾

B⁺ + H₂O  BOH + H⁺

b) Degree of Hydrolysis

c) pH

B⁺ + H₂O  BOH + H⁺

c                         0         0         original conc

c(1-h)                 ch        ch      conc at equilibrium

  = ch = (K_w /K_b.c ) ^½= (K_w .c /K_b)^½ = (K_w .c /K_b )^½

pH = – log  

pH = -log (K_w .c /K_b )^½

pH = -½ ( log K_w -log K_b + log c )

pH = ½ ( pK_w -pK_b – log c)

pH = 7 – ½(pK_b + log c)

 

Salts of weak Acid and weak base

a) Hydrolysis constant

K_h = K_w / K_a .K_b

b) Degree of Hydrolysis

c) pH

HA H⁺ + A‾

 = K_a (HA / A‾)

  = K_a (ch / c(1-h))

 = K_a (h /1-h)

√K_h = (h /1-h)

  = K_a √K_h =K_a  (K_w/ K_a .K_b)^½ = (K_a .K_w /K_b)^½

pH = -log 

pH = – log (K_a .K_w /K_b)^½

pH = -½( log K_a + log K_w -log K_b)

pH =½ (pK_w +pK_a -pK_b)

pH = 7 + ½ (pK_a – pK_b)

If pK_a < pK_b , pH of the solution will be less than 7 and the solution will be acidic.

If pK_a > pK_b , pH of the solution will be greater than 7 and the solution will be basic.

If pK_a = pK_b , pH of the solution will be equal to 7 and the solution will be neutral.