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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
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
Ka of HCN = 4.8 × 10–9
Hence , Kb of its conjugate base `"CN"^(-)`
`= "K"_"w"/"K"_("a")`
`= 10^(-14)/(4.8 xx 10^(-9))`
`= 2.08 xx 10^(-6)`
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