If K_a for HCN is 6.17 $\times$ 10^-¹⁰,what is K_b for CN^-? Note: CN^- + H₂O HCN + OH^- A)6.17 $\times$ 10^-²⁴ B)6.17 $\times$ 10⁴ C)1.62 $\times$ 10^-⁵ D)1.23 $\times$ 10^-⁹ E)none of these

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Quizplus · Accepted Answer

To solve this problem, we need to use the equilibrium constant expression for the reaction given in the note. HCN + H₂O ⇌ CN^- + OH^- The equilibrium constant expression for this reaction is: KC = [CN^-][OH^-]/[HCN][H₂O] We are given the value of the equilibrium constant (K_a) for this reaction in terms of standard free energy change (ΔG°) as: K_a = e^(-ΔG°/RT) We are asked to find the equilibrium constant (K_b) for the reaction of HCN with H₂O to form CN^- and OH^-. This reaction can be obtained by reversing the reaction given in the note. To obtain the equilibrium constant expression for this reaction, we need to reverse the direction of the equilibrium constant expression for the reaction given in the note: KC' = [HCN][H₂O]/[CN^-][OH^-] The equilibrium constant for the reverse reaction (KC') is related to the equilibrium constant for the forward reaction (KC) by the equation: KC' = 1/KC Substituting the value of K_a into the equation K_a = e^(-ΔG°/RT), we get: K_a = e^(-ΔG°/RT) = KC Taking the reciprocal of both sides, we get: 1/K_a = 1/KC' Substituting the values given, we get: 1/(6.17 F1F1F1S1F1F1F10 10^-²) = KC' Simplifying this expression, we get: KC' = 1.62 F1F1F1S1F1F1F10 10^-⁵ Therefore, the answer is option C.

If Ka for HCN Is 6 ×\times× 10-10,what Is Kb for CN-? Note: CN- + H2O

If K_a for HCN Is 6 $\times$ 10^-¹⁰,what Is K_b for CN^-? Note: CN^- + H₂O