Question

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

Answer 1

Answer:

10.86

Explanation:

Given that:

$K_{b}=1.8\times 10^{-5}$

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:

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

$1.8\times 10^{-5}=\frac {x^2}{0.0288-x}$

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