The conjugate base of $HBr$ is ......
- A$H_2Br^{+}$
- B$H^{+}$
- C$Br^{-}$
- D$Br^{+}$
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For any diprotic acid of the type $H_2A$,the relationship between the first $(Ka_1)$ and second $(Ka_2)$ ionization constants is:
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Among the following,the dissociation constant $(K_a)$ is highest for:
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View SolutionGiven:
$(i) \ HCN_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + CN^{-}_{(aq)}$
$K_a = 6.2 \times 10^{-10}$
$(ii) \ CN^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons HCN_{(aq)} + OH^{-}_{(aq)}$
$K_b = 1.6 \times 10^{-5}$
These equilibria show the following order of the relative base strength:
$(i) \ HCN_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + CN^{-}_{(aq)}$
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$(ii) \ CN^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons HCN_{(aq)} + OH^{-}_{(aq)}$
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These equilibria show the following order of the relative base strength:
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