Question

What is the pH of a 0.0288 M solution of ammonia (Kb 1.8 x 10-5)?

Answer 1

10.86

Step-by-step explanation:

Given that:

`K_(b)=1.8* 10^(-5)`

Concentration = 0.0288 M

Consider the ICE take for the dissociation of ammonia as:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

At t=0 0.0288 - -

At t =equilibrium (0.0288-x) x x

The expression for dissociation constant of ammonia is:

`K_(b)=\frac {\left [ NH_4^(+) \right ]\left [ {OH}^- \right ]}{[NH_3]}`

`1.8* 10^(-5)=\frac {x^2}{0.0288-x}`

x is very small, so (0.0288 - x) ≅ 0.0288

Solving for x, we get:

x = 7.2×10⁻⁴ M

pOH = -log[OH⁻] = -log(7.2×10⁻⁴) = 3.14

Also,

pH + pOH = 14

So, pH = 10.86