Salt hydrolysis
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Salt hydrolysis-
Q 1) Calculate for 0.01 N solution of CH3COONa
i) hydrolysis constant (ii) degree of hydrolysis (iii) pH
(Ka = 1.9 x 10-5)
Solution )
i)
Sodium acetate is a salt of strong base and weak acid.
Kw = 10-14
Kh =Kw / Ka
Kh = 10-14 /1.9 x 10-5
Hydrolysis constant ‘Kh ‘= 5.26 x 10 -10 Ans.
ii)
Salt hydrolysis-
CH3COO– + H2O ⇌ CH3COOH + OH–
C (1-h) Ch
C = 0.01, Kh = 5.26 x 10 -10
Kh = Ch2
h = √ Kh / C
h = √ 5.26 x 10 -10 / 0.01
h = √ 526 x 10 -10
h = 22.9 x 10 -5
Degree of hydrolysis, h = 2.29 x 10-4 Ans.
iii)
[OH–] = Ch = 0.01 x 2.29 x 10 -4 = 2.29 x 10-6 M
[H+] = Kw / [OH–] = 10-14 / 2.29 x 10-6
[H+] = 4.37 x 10 -9 M
pH = -log [H+]= – log (4.37 x 10 -9)
= – (log 4.37 + log 10-9)
= – (log 4.37 -9 log 10)
= -log 4.37 + 9 log 10
= -0.6405 + 9= 8.36
pH = 8.36 Ans.
Alternate method for calculation of pH-
CH3COONa is a salt of strong base and weak acid. Therefore ,
pH = 7 + 1/2 pKa + 1/2 log C
Ka =1.9 x 10 -5, C = 0.01 M
pKa = -log Ka= -log (1.9 x 10 -5)
= -(log 1.9 + log 10 -5)
= -(log 1.9 – 5 log 10 )
= – ( 0.2787 – 5 x 1)
= – ( – 4.7213)
pKa = 4.7213
pH = 7 + (4.7213/2) + 1/2 log 0.01=7 + 2.36 + (-2/2)
pH= 9.36 -1.00
pH = 8.36 Ans.
Q 2) Calculate the % hydrolysis in 0.003 M aqueous solution of NaOCN . Ka for HOCN = 3.33 x 10 -4.
Solution )
It is a salt of strong base and weak acid.
h = √ Kw/(Ka. C)
C = 0.003 M , Ka = 3.33 x 10 -4 , Kw = 10 -14
h = √ 10 -14 / 3.33 x 10-4 x 0.003
h = √ 10 -8
h = 10 -4
% of h = 10 -4 x 100 = 10-2
% of h = 0.01 % Ans.
Q 3) Calculate the pH of a 0.5 M aqueous NaCN solution . Given pKb 0f CN – = 4.7
Solution )
pKa for HCN = 14 – 4.7= 9.3
NaCN is a salt of strong base and weak acid.Therefore,
[OH–] = Ch
h = √ Kw/(Ka. C)
[OH–] = C √ Kw/(Ka. C)
[OH–] = √ Kw .C / Ka
Taking log and reverting the sign-
-log [OH–] = 1/2 [-log Kw – log C + log Ka]
-log [OH–] = pOH , -log Kw = pKw , -log Ka = pKa
pOH = 1/2 [pKw -log C – pKa]
= 1/2[14 – log 0.5 – 9.3]
= 1/2 [ 4.7 + 0.3010]
pOH = 2.5
pH = 14 – pOH = 14 – 2.5