What is the value of K_b for the cyanide anion, CN¯? K_a(HCN) = 6.2 * 10¯¹⁰

Question

Quizplus · Accepted Answer

Using the given values: K_a(HCN) = 6.2 F1F1F1S1F1F1F10 10¯¹⁰ We can write the reaction as: HCN + H2O ⇌ CN¯ + H3O+ where K_a(HCN) = [H3O+][CN¯]/[HCN][H2O] For the cyanide anion, we need to find K_b which is given as: K_a(HCN) = K_b/([OH¯][CN¯]) We know that in water, Kw = [H3O+][OH¯] = 1.0 × 10¯14. At 25°C, [OH¯] = Kw/[H3O+] = 1.0 × 10¯14/10¯pH = 10^-(14+pH) Substituting the values: 6.2 F1F1F1S1F1F1F10 10¯¹⁰ = K_b/([OH¯][CN¯]) 6.2 F1F1F1S1F1F1F10 10¯¹⁰ = K_b/[10^-(14+pH)][CN¯] K_b = (6.2 F1F1F1S1F1F1F10 10¯¹⁰)(10^-(14+pH)) Now, for CN¯, K_a(HCN) = [H3O+][CN¯]/[HCN][OH¯] Assuming [OH¯] = [H3O+] 6.2 F1F1F1S1F1F1F10 10¯¹⁰ = [H3O+]^2/[CN¯] = 10^-pH/([CN¯]) Solving for [CN¯]: [CN¯] = 10^(-pH) / (6.2 F1F1F1S1F1F1F10 10¯¹⁰) Substituting this value in the expression for K_b, we get: K_b = (6.2 F1F1F1S1F1F1F10 10¯¹⁰)(10^-(14+pH))/(10^(-pH)/6.2 F1F1F1S1F1F1F10 10¯¹⁰) Simplifying, we can obtain: K_b = 1.6 F1F1F1S1F1F1F10 10¯¹⁵ Therefore, the correct answer is B) 1.6 F1F1F1S1F1F1F10 10¯¹⁵.

What Is the Value of Kb for the Cyanide Anion

What Is the Value of K_b for the Cyanide Anion