Question

The $ \;pH $ of an aqueous solution of ammonia is $ \;11.5 $ . Find the molarity of the solution. [Given that: $ {{K}_{b(N{{H}_{4}}OH)}}=1.8\times {{10}^{-5}} $ ]
(A) $ 0.556\;M $
(B) $ 0.66\;M $
(C) $ 5.6\;M $
(D) $ 6.6\;M $

Answer 1

Hint: Ammonia is always present in aqueous solution as Ammonium Hydroxide. Hence, from the given value of $ \;pH $ we can find the concentration of $ H^+ $ and $ OH^- $ . Substituting the obtained values and the values obtained from ICE (Initial Change Equilibrium) in the equation of ionisation constant, and find the concentration of Ammonium Hydroxide.

Complete answer:
The chemical formula for aqueous solution of ammonia is $ N{{H}_{4}}OH $
Here, the $ pH\; $ of the solution is given as $ pH=11.5 $ and $ pH\; $ can be expressed in terms of concentration of Hydrogen ion as
 $ pH=-{{\log }_{10}}\left[ {{H}^{+}} \right] $
 $ \therefore 11.5=-{{\log }_{10}}\left[ {{H}^{+}} \right] $
Changing the logarithm to exponential,
 $ \therefore \left[ {{H}^{+}} \right]={{10}^{-11.5}} $
Now, to find the concentration of hydroxyl ion, we can use the total concentration of Hydrogen ion and Hydroxyl ion in neutral solution, which is expressed as
 $ \left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}} $
 $ \therefore {{10}^{-11.5}}\left[ O{{H}^{-}} \right]={{10}^{-14}} $
Simplifying the equation further,
 $ \therefore \left[ O{{H}^{-}} \right]=\dfrac{{{10}^{-14}}}{{{10}^{-11.5}}} $
 $ \therefore \left[ O{{H}^{-}} \right]={{10}^{-2.5}} $
Now, the dissociation equation of the base can be expressed as
 $ N{{H}_{4}}O{{H}_{(aq)}}\rightleftharpoons NH_{4(aq)}^{+}+O{{H}^{-}}_{(aq)} $
From this reaction, the ionisation constant $ K_b $ is calculated as
 $ {{K}_{b}}=\dfrac{\left[ N{{H}_{4}}^{+} \right]\left[ O{{H}^{-}} \right]}{\left[ N{{H}_{4}}OH \right]} $
From the dissociation reaction, we can understand that for every ion dissociated, we get the same number of Ammonium ions and Hydroxyl ions. Thus, the concentration of both ions will be equal.
Now, we have obtained the values
 $ {{K}_{b}}=1.8\times {{10}^{-5}} $ , $ \left[ N{{H}_{4}}^{+} \right]=\left[ O{{H}^{-}} \right]={{10}^{-2.5}} $
Substituting these values,
 $ 1.8\times {{10}^{-5}}=\dfrac{{{\left( {{10}^{-2.5}} \right)}^{2}}}{\left[ N{{H}_{4}}OH \right]} $
 $ \left[ N{{H}_{4}}OH \right]=\dfrac{{{\left( {{10}^{-2.5}} \right)}^{2}}}{1.8\times {{10}^{-5}}} $
Simplifying the powers,
 $ \left[ N{{H}_{4}}OH \right]=\dfrac{{{10}^{-5}}}{1.8\times {{10}^{-5}}} $
 $ \therefore \left[ N{{H}_{4}}OH \right]=0.556M $
As the concentration is calculated in Molarity, this is the molarity of ammonium hydroxide.
Hence, the correct answer is Option $ (A) $ .

Note:
The point to remember is that ammonia on dissolution in water will always convert to ammonium hydroxide which is a base. The concentration of $ OH^- $ can also be found from $ pOH $ where $ pH+pOH=14 $. Whenever we are given an ionisation of weak acid or weak base, the concentration of ions obtained by dissociation will depend on each other or say can be defined by a single variable.