Question

What is the pH of a 1.5 M solution of ammonia? The dissociation constant of ammonia at ${{25}^{\circ }}C$ is $1.80\times \,{{10}^{-5}}$.

Answer 1

Hint: Before answering this question, we should know about the base dissociation constant. To measure the basicity of a base or its strength, we use the Base dissociation constant. We use the expression $p{{K}_{a}}+p{{K}_{b}}=14$to calculate it.

Complete answer:
Ammonia is a weak base in an aqueous solution, so as we know the pH range of the base is less than 7.
$N{{H}_{3}}(aq)+{{H}_{2}}O(l)\to NH_{4}^{+}(aq)+O{{H}^{-}}(aq)$
Base dissociation constant (${{K}_{b}}$) gives the ratio between the equilibrium concentrations of the ammonium constants and of the hydroxide anions and the equilibrium concentrations of ammonia.
${{K}_{b}}=\dfrac{[N{{H}_{4}}^{+}].\,[O{{H}^{-}}]}{[N{{H}_{3}}]}$
There will be a production of ammonium cations and hydroxide anions by partial ionization. The concentration of ammonia that ionizes can be taken as xM, At equilibrium, the solution will have
[$NH_{4}^{+}$]= $[O{{H}^{-}}]$= xM
1 mole of ammonium cation and 1 mole of hydroxide ions are produced by every mole of ammonia.
If xM is ionizing, it can be expected that the solution has xM of the two ions.
The solution will now have
$[N{{H}_{3}}]=(1.5-x)\,M$
The initial concentration of ammonia will decrease by xM when xM will ionize.
Now, the expression of the base dissociation constant will be
${{K}_{b}}=\dfrac{{{x}^{2}}}{1.5-x}$
$1.80\times {{10}^{-5}}=\dfrac{{{x}^{2}}}{1.5-x}$
The value of the initial concentration of the base is greater than the value of the base dissociation constant. Therefore,
$1.5-x\approx 1.5$
The value of the initial concentration of ammonia is greater than the value of the base dissociation constant. That is why the ionization equilibrium will be on the left.
$1.80\times {{10}^{-5}}=\dfrac{{{x}^{2}}}{1.5}$
Now, to find the value of x
$x=\sqrt{1.5\times 1.8\times {{10}^{-5}}}=0.005196$
The equilibrium concentration of the hydroxide anions is represented by xM,
$[O{{H}^{-}}]$= 0.005196 M
At ${{25}^{\circ }}C$, an aqueous solution has-
$pH+pOH=14$
As
$pOH=-\log (O{{H}^{-}})$
So, pH = 14 +log($O{{H}^{-}}$)
Put the values in the expression :
pH= 14+log(0.005196)
     = 11.72

Note:
pH has various uses like
Hydrochloric acid in our stomach helps to digest the food without damaging our stomach. But if the concentration of acid exceeds a certain limit, then we face problems like pain and indigestion. So, to neutralize it Antacids are taken to balance the pH.
Soils need a certain amount of pH to be fertile. The soil should not be too acidic or too alkaline. It hampers the growth of the plant.