Question

What is the equilibrium concentration of ammonium ion in a 0.33 M solution of ammonia (NH3, Kb = 1.8 × 10–5) at 25oC?

An initially 1.0 M aqueous solution of a weak monoprotic acid has a total ion concentration of M when equilibrium is established.

What is the acid-ionization constant, Ka, of the weak acid? (assume Ca/Ka ≥ 102) For the reaction ΔG°700K = –13.457 kJ. What is Kp for this reaction at 700. K?

Answer 1

a)  NH₃ + H₂O <-----> NH₄⁺ + OH^-
Here from ICE table we will have, [NH₄⁺] = x , [OH^-] =x , [NH₃] =0.33 M - x
K_b = [NH₄⁺][OH^-] / [NH₃]
=> 1.8 x 10^-5 = x * x / (0.33 M - x)

=> 1.8 x 10^-5 = x² /( 0.33 - x)
Let us assume that x << 0.33 M , so 0.33 - x =0.33

Now, 1.8 x 10^-5 = x² / 0.33

=> 0.594 x 10^-5 = x²

=> x² = 5.94 x 10^-6

=> x = 5.94 x 10^-6

=> x = 2.44 x 10^-3 M
The equilibrium concentration of ammonium ion, [NH₄⁺] = x = 2.44 x 10^-3 M.

b) HA H⁺ + A^-

Total ion concentration is not mentioned, it is simply given as M.

so, [H⁺] = [A^-] = 1/2 * M(total ion concentration) = 0.5 M

Now, K_a = [H⁺] [A^-] / [HA]

=> K_a = (0.5 )² / 1.0

=> K_a = (0.25 /1.0)

=> K_a = 0.25

c) Given, G₀ at 700 K = -13.457 KJ

We have the relation as ,

G₀ = - RT ln K_p

=> -13.457 KJ/mol = -8.314 J/mol K x 700 K x ln K_p

=> -13457 J/mol = = -5819.8 J/mol x lnK_p

=> (-13457 /-5819.8) = lnK_p

=> ln K_p = 2.3123

=> K_p = e^(2.3123)

=> K_p = 10.097