Question

What is the equilibrium constant for the reaction of NH3 with water?

Answer 1

The equilibrium constant for the reaction of NH₃ with water is 1.76 × 10⁻⁵.

In aqueous solution, ammonia acts as a base. It accepts hydrogen ions from H₂O to yield ammonium and hydroxide ions.

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

The base ionization constant is

`K_"b" = (["NH"_4^+]["OH"^-])/(["NH"_3])`

We can determine the `K_"b"` value from pH measurements.

Example

The pH of a 0.100 mol/L solution of NH₃ is 11.12. What is the `K_"b"` for NH₃?

Solution

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

pH = 11.12

pOH = 14.00 – 11.12 = 2.88

[OH⁻] = `10^"-pOH" = 10^-2.88` = 1.32 × 10⁻³ mol/L
[NH₄⁺]= 1.32 × 10⁻³ mol/L
[NH₃] = (0.100 -1.32 × 10⁻³) mol/L = 0.099 mol/L

`K_"b" = (["NH"_4^+]["OH"^-])/(["NH"_3])` = (1.3×10⁻³ × 1.32×10⁻³)/0.099 = 1.76 × 10⁻⁵