Question

calculate the concentration of NH3 (pKa (NH4+/NH3)= 9.25) if the pH =10. please explain how to get to the answer (0.66mM)

Answer 1

Given, pH of NH₃ = 10

Therefore, pOH = 14-10 = 4

pOH = -log[OH^-] = 4

[OH^-] = 10^-4 = 0.0001 M

pKb = 14- pKa = 14- 9.25 = 4.75

Kb = 10^-4.75 = 1.78 x 10^-5

Now, let us assume the initial concentration of [NH₃] = x M

Constructing the ICE table

	NH3(aq)	H2O(l)		NH4+(aq)	OH-(aq)
Initial	x M			0	0
Change	-0.0001 M			+ 0.0001M	+0.0001 M
Equilibrium	x - 0.0001 M			0.0001 M	0.0001 M

Kb = [NH₄⁺][ OH^-]/[ NH₃]

1.78 x 10^-5 = (0.0001)²/(x - 0.0001)

(x - 0.0001) = (0.0001)²/(1.78 x 10^-5) = 0.000562 M

x = 0.000562 + 0.0001 M = 0.00066 M

x = [NH₃] = 0.00066 M = 0.66 mM

Concentration of NH₃ = 0.66 mM