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