InterviewSolution
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InterviewSolution
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The ionization constant of HF, HCOOH and HCN at 298 K are `6.8xx10^(-4), 1.8xx10^(-4) and 4.8xx106(-9)` respectively. Calculate the ionization constant of the corresponding conjugate base. |
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Answer» For `F^(-) , " " K_(b) = K_(w)//K_(a) = 10^(-14)//(6.8xx10^(-4))=1.47 xx 10^(-11) ~= 1.5 xx 10^(-11)`. For `HCO O^(-) , " " K_(b) = 10^(-14)//(1.8xx10^(-4))=5.6xx10^(-11)` For `CN^(-), " " K_(b) = 10^(-14) // (4.8xx10^(-9))=2.08 xx 10^(-6)` |
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