Question

The equilibrium constant, K. for a neutralization reaction can be symbolized as K_n. Using K_w, and the K_a and K_b values from these tables, determine K_n for the following reaction at 25 degree C. NH_3 (aq) + HCOOH(aq) NH_4^+ (aq) + COOH^- (aq)

Answer 1

Let's write the equations for Kn, Ka, Kb and Kw.

Let's write the equations for Kn, Ka, Kb and Kw. K_n = [NH^+_4][COOH^-]/[NH_3][HCOOH] K_a = [H^+][COOH^-]/[HCOOH] K_b = [NH^+_4][OH^-]/[NH_3][H_2O] K_w = [H^+][OH^-]/[H_2O] Therefore, by substituting each of the above terms into our equation for Kn, we can get the following. K_n = K_bK_a/K_w K_n = 1.8 * 10^-5 * 1.8 * 10^-4/1 * 10^-14 = 3.24 * 10^5

$K_{a}=\frac{[H^{+}][COOH^{-}]}{[HCOOH]}$

$K_{b}=\frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}][H_{2}O]}$

$K_{w}=\frac{[H^{+}][OH^{-}]}{[H_{2}O]}$

Therefore, by substituting each of the above terms into our equation for Kn, we can get the following.

$K_{n}=\frac{K_{b}K_{a}}{K_{w}}$

$K_{n}=\frac{1.8*10^{-5}*1.8*10^{-4}}{1*10^{-14}}=3.24*10^{5}$