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The Ionization Constant of Hf, Hcooh and Hcn at 298k Are 6.8 × 10–4, 1.8 × 10–4 And 4.8 × 10–9 Respectively. Calculate the Ionization Constants of the Corresponding Conjugate Base - CBSE (Science) Class 11 - Chemistry

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Question

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Solution 1

It is known that,

`K_b = K_w/K_a`

Given,

Ka of HF = 6.8 × 10–4

Hence, Kb of its conjugate base F

= `K_w/K_a`

`= 10^(-14)/(6.8 xx 10^(-4))`

`= 1.5 xx 10^(-11)`

Given,

Ka of HCOOH = 1.8 × 10–4

Hence, Kb of its conjugate base HCOO

= `K_w/K_a`

`= 10^(-14)/(1.8 xx 10^(-4))`

`= 5.6 xx 10^(-11)`

Given

`K_a of HCN = 4.8 xx 10^(-9)`

Hence , K_b of its conjugate base `CN^(-)`

`= K_w/K_(a)`

`= 10^(-14)/(4.8 xx 10^(-9))`

`= 2.08 xx 10^(-6)`

Solution 2

For F– , Kb =Kw/Ka= 10-14/(6.8 x 10-4) = 1.47 x 10-11 = 1.5 x 10-11 .

For HCOO-, Kb = 10-14/(1.8 x 10-4) = 5.6 x 10-11 

For CN, Kb= 10-14/(4.8 X 10-9) = 2.08 x 10-6

  Is there an error in this question or solution?

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Solution The Ionization Constant of Hf, Hcooh and Hcn at 298k Are 6.8 × 10–4, 1.8 × 10–4 And 4.8 × 10–9 Respectively. Calculate the Ionization Constants of the Corresponding Conjugate Base Concept: Ionization of Acids and Bases - Ionization Constants of Weak Acids.
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