Given (i) HCN(aq) + H2O( l) leftharpoons H3O+(aq) + CN- (aq)Ka =6.2 × 10-10 (ii) CN (aq) + H2O(l) leftharpoons HCN (aq) + OH- (aq) Kb= 1.6 × 10-5. These equilibria show the following order of the relative base strength,

Question

Given (i) HCN(aq) + H2O( l) leftharpoons H3O+(aq) + CN- (aq)Ka =6.2 × 10-10 (ii) CN (aq) + H2O(l) leftharpoons HCN (aq) + OH- (aq) Kb= 1.6 × 10-5. These equilibria show the following order of the relative base strength, (A) OH- > H2O > CN- (B) OH- > CN- > H2O (C)  H2O > CN- > OH- (D)  CN- > H2O > OH-

Tardigrade Admin · Accepted Answer

The more is the value of equilibrium constant, the more is the completion of reaction or more is the concentration of products i.e. the order of relative strength would be OH- > CN- > H2O

Q. Given
$(i) H CN (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + C N^{-} (a q) K_{a} = 6.2 \times 1 0^{- 10}$
$(ii) CN (a q) + H_{2} O (l) ⇌ H CN (a q) + O H^{-} (a q) K_{b} = 1.6 \times 1 0^{- 5} .$
These equilibria show the following order of the relative base strength,

$O H^{-} > H_{2} O > C N^{-}$

$O H^{-} > C N^{-} > H_{2} O$

$H_{2} O > C N^{-} > O H^{-}$

$C N^{-} > H_{2} O > O H^{-}$

The more is the value of equilibrium constant, the more is the completion of reaction or more is the concentration of products i.e. the order of relative strength would be
$O H^{-} > C N^{-} > H_{2} O$

Q. Given (i)HCN(aq)+H2​O(l)⇌H3​O+(aq)+CN−(aq)Ka​=6.2×10−10 (ii)CN(aq)+H2​O(l)⇌HCN(aq)+OH−(aq)Kb​=1.6×10−5. These equilibria show the following order of the relative base strength,

OH−>H2​O>CN−

OH−>CN−>H2​O

H2​O>CN−>OH−

CN−>H2​O>OH−