The K_afor HCN is 4.9 × 10^-¹⁰. What is the value of K_bfor CN^-?

Question

Quizplus · Accepted Answer

To find the value of K_bfor CN^-, we need to use the principle of inverse solubility products. The inverse solubility product of HCN is given as: K_afor HCN = 4.9 × 10^-¹⁰ We can write the balanced equation for the dissociation of HCN and the formation of CN- as follows: HCN ⇌ H+ + CN- The equilibrium constant expression for this reaction is: K_a= [H+][CN-]/[HCN] We can use the stoichiometry of the balanced equation to relate the concentrations of H+ and CN-: [H+] = [HCN] - [CN-] Substituting this expression into the equilibrium constant expression gives: K_a= ([HCN]-[CN-])[CN-]/[HCN] Simplifying, [CN-]2 / [HCN] = K_a/([HCN]-[CN-]) We are given K_afor HCN and asked to find K_bfor CN^-. Note that CN- and CN^- differ only by the central atom. Since they have the same charge and similar size, we can assume that their solubility products are similar. Thus: K_bfor CN^- ≈ K_afor CN- To find [CN-] for a given concentration of HCN, we need to solve the quadratic equation obtained by substituting the given values into the above equation: Solving for [CN-], we get: [CN-] = (K_a/[HCN]) + [(K_a/[HCN])2 + 4K_a]/2 Since we know that [CN-] + [HCN] = [CN^-] and [HCN] is given, we can calculate [CN^-] using the calculated value of [CN-]. This gives: K_bfor CN^- ≈ 2.0 × 10^-⁵. Thus, the correct answer is B.

The Ka for HCN Is 4

The K_afor HCN Is 4