Salt hydrolysis

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Salt hydrolysis-

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

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

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

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

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