question_answer 71)

The ionisation constant of HF, HCOOH and HCN at 298 K are \[6.8\text{ }\times \text{ }{{10}^{-4}},\text{ }1.8\text{ }\times \text{ }{{10}^{-4}}\] and\[4.8\times {{10}^{-9}}\]respectively. Calculate the ionisation constants of the corresponding conjugate base.

Answer:

If \[{{K}_{a}}\] and \[{{K}_{b}}\] are the ionisation constants of acids and their conjugate bases, then \[{{K}_{a}}\times {{K}_{b}}={{K}_{w}}\] \[\mathbf{HCN:}\] \[{{K}_{a}}=4.8\times {{10}^{-9}}\] \[\therefore \] \[4.8\times {{10}^{-9}}\times {{K}_{b}}={{10}^{-14}}\] \[{{K}_{b}}=2.083\times {{10}^{-6}}\] \[\mathbf{HCOOH:}\] \[{{K}_{a}}=1.8\times {{10}^{-4}}\] \[\therefore \] \[1.8\times {{10}^{-4}}\times {{K}_{b}}={{10}^{-14}}\] \[{{K}_{b}}=5.55\times {{10}^{-11}}\] \[\mathbf{HF:}\] \[{{K}_{a}}=6.8\times {{10}^{-14}}\] \[\therefore \] \[6.8\times {{10}^{-14}}\times {{K}_{b}}={{10}^{-14}}\] \[{{K}_{b}}=1.47\times {{10}^{-11}}\]