Salt hydrolysis Numerical part 1

Salt hydrolysis

source: Worksheets Library

Salt hydrolysis-

Q 1) Calculate the degree of hydrolysis and pH of 0.1 M sodium acetate solution. Hydrolysis constant of CH₃COONa is 5.6 x 10^-10.

Solution )

Sodium acetate is a salt of weak acid and strong base.

On Salt hydrolysis-

                                   CH₃COO^–   +     H₂O     ⇌       CH₃COOH      + OH-

                                     weak ion   

Before hydrolysis      C                                                 0                         0

After hydrolysis      C ( 1-h)                                          Ch                   Ch

K_h = [CH₃COOH][OH^–] / [CH₃COO^–]

K_h  = Ch x Ch / C (1-h)

K_h = Ch²/(1-h)

because,   h <<<1 ,

So,

1-h =1

K_h = Ch2

C = 0.1 M , Kh = 5.26 x 10 ^-10

h = √ K_h / C

h = √ 5.6 x 10 ^-10  / 0.1

h = √ 56 x 10 ^-10

Degree of hydrolysis ‘h’  = 7.48 x 10 ^-5 Ans.

[OH^–] = Ch =0.1 x 7.48 x 10 ^-5 = 7.48 x 10 ^-6 M

[H⁺ ] [OH^–] = K_w

[H⁺] = K_w / [OH^–] = 10 ^-14 / 7.48 x 10 ^-6

[H⁺] = 1.33 x 10 ^-9 M

pH = -log [H⁺]

= – log [1.33 x 10 ^-9]

= – [log 1.33 – 9 log 10]

= – log 1.33 + 9 log 10]

= – 0.124 +9=8.876

pH = 8.9 Ans.

Q 2) Calculate the hydrolysis constant for NH₄Cl , pH and [OH^–] in 0.1 M NH₄Cl solution ?

(K_b or K _NH4OH = 1.75  x 10^-5 , K_w = 1 x 10^-14)

Solution )

i)

K_h = K_w / K_b

Given, K_b= 1.75  x 10^-5 , K_w = 1 x 10^-14

K_h = 1 x 10^-14 / 1.75  x 10^-5

Hydrolysis constant ‘K_h‘ = 5.7 x 10 ^-10 Ans.

ii) Calculation for pH-

On Salt hydrolysis-

                                          NH₄ ⁺    +      H₂O     ⇌     NH₄OH  +  H ⁺

                                    weak ion  

Before hydrolysis      C                                                 0                   0

After hydrolysis      C ( 1-h)                                         Ch                 Ch

K_h = Ch x Ch /C ( 1-h) = Ch² /(1-h)

h <<<1

So, (1-h) = 1

K_h = Ch²

C = 0.1 M

h = √ K_h / C

h = √ 5.7 x 10 ^-10  / 0.1

h = √ 57 x 10 ^-10

h = 7.55 x 10 ^-5 Ans.

[H⁺] = Ch = 0.1 x 7.55 x 10 ^-5

[H⁺ ] = 7.55 x 10 ^-6

pH = -log [H⁺] = – log (7.55 x 10 ^-6)

= – (log 7.55 – 6 log 10)

= – ( 0.8779 – 6 x 1)

= -0.8779 + 6

pH = 5.12 Ans.

Alternate method for calculation of pH-

NH₄Cl is a salt of strong acid and weak base . Therefore ,

pH = 7 – 1/2 pK_b – 1/2 log C

K_b =1.75 x 10 ^-5, C = 0.1 M

pK_b = -log K_b= -log (1.75 x 10 ^-5)

= -(log 1.75 + log 10 ^-5)

= -(log 1.75 – 5 log 10 )

= – ( 0.245 – 5 x 1)

= – ( – 4.755)

pK_b = 4.755

pH = 7 – (4.755/2) – 1/2 log 0.1

=7 –  2.377 – (-1/2)

= 7.5 – 2.377

pH = 5.123 Ans.

iii) Calculation for [OH^–]

[H⁺][OH^–] = 10^-14

[OH^–] = 10 ^-14 / 7.55 x 10^-6

[OH^–] = 1.32 x 10^-9 M Ans.

Q 3) Calculate the pH of an aqueous solution of 1.0 M  ammonium formate

( HCOONH₄) assuming complete dissociation . pK_a  for HCOOH = 3.8 , pK_b for NH₄OH = 4.8

Solution )

Ammonium formate is a salt of weak acid and weak base.So,

pK_a = 3.8 , pK_b = 4.8

pH = 7 + 1/2 pK_a – 1/2 pK_b

= 7 + (3.8/2) – (4.8/2)

= 7 + 1.9 – 2.4 = 6.5

pH = 6.5 Ans.