Question

calculate Ka for 0.10 M NH4Cl (pH= 4.86) ?

Answer 1

pH = 4.86

pH = -log [H⁺]

4.86 = -log [H⁺]

[H⁺] = 1.38 $\times$ 10^-5 M

NH₄⁺$\rightarrow$ NH₃ + H⁺
The equilibrium constant for this weak acid is:
K_a = [NH₃] [H⁺] / [NH₄⁺]

Using ICE, we would get for the equilibrium (i.e. final) concentrations (let x = [H⁺]):

[NH₄⁺] = 0.1 - x (initial 0.1 M, change -x)
[NH₃] = x (initial 0, change + x)
[H⁺] = x (initial 0, change + x)

K_a = x² / 0.1-x

0.1>>>x, so we can write 0.1 instead of 0.1-x

K_a = x² / 0.1 = (1.38 $\times$ 10^-5)²/0.1 = 1.90 $\times$ 10^-9