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Question: The \( \;pH \) of an aqueous solution of ammonia is \( \;11.5 \) . Find the molarity of the solution...

The   pH\;pH of an aqueous solution of ammonia is   11.5\;11.5 . Find the molarity of the solution. [Given that: Kb(NH4OH)=1.8×10−5{{K}_{b(N{{H}_{4}}OH)}}=1.8\times {{10}^{-5}} ]
(A) 0.556  M0.556\;M
(B) 0.66  M0.66\;M
(C) 5.6  M5.6\;M
(D) 6.6  M6.6\;M

Explanation

Solution

Ammonia is always present in aqueous solution as Ammonium Hydroxide. Hence, from the given value of   pH\;pH we can find the concentration of H+H^+ and OH−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 NH4OHN{{H}_{4}}OH
Here, the pH  pH\; of the solution is given as pH=11.5pH=11.5 and pH  pH\; can be expressed in terms of concentration of Hydrogen ion as
pH=−log⁡10[H+]pH=-{{\log }_{10}}\left[ {{H}^{+}} \right]
∴11.5=−log⁡10[H+]\therefore 11.5=-{{\log }_{10}}\left[ {{H}^{+}} \right]
Changing the logarithm to exponential,
∴[H+]=10−11.5\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
[H+][OH−]=10−14\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}
∴10−11.5[OH−]=10−14\therefore {{10}^{-11.5}}\left[ O{{H}^{-}} \right]={{10}^{-14}}
Simplifying the equation further,
∴[OH−]=10−1410−11.5\therefore \left[ O{{H}^{-}} \right]=\dfrac{{{10}^{-14}}}{{{10}^{-11.5}}}
∴[OH−]=10−2.5\therefore \left[ O{{H}^{-}} \right]={{10}^{-2.5}}
Now, the dissociation equation of the base can be expressed as
NH4OH(aq)⇌NH4(aq)++OH−(aq)N{{H}_{4}}O{{H}_{(aq)}}\rightleftharpoons NH_{4(aq)}^{+}+O{{H}^{-}}_{(aq)}
From this reaction, the ionisation constant KbK_b is calculated as
Kb=[NH4+][OH−][NH4OH]{{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
Kb=1.8×10−5{{K}_{b}}=1.8\times {{10}^{-5}} , [NH4+]=[OH−]=10−2.5\left[ N{{H}_{4}}^{+} \right]=\left[ O{{H}^{-}} \right]={{10}^{-2.5}}
Substituting these values,
1.8×10−5=(10−2.5)2[NH4OH]1.8\times {{10}^{-5}}=\dfrac{{{\left( {{10}^{-2.5}} \right)}^{2}}}{\left[ N{{H}_{4}}OH \right]}
[NH4OH]=(10−2.5)21.8×10−5\left[ N{{H}_{4}}OH \right]=\dfrac{{{\left( {{10}^{-2.5}} \right)}^{2}}}{1.8\times {{10}^{-5}}}
Simplifying the powers,
[NH4OH]=10−51.8×10−5\left[ N{{H}_{4}}OH \right]=\dfrac{{{10}^{-5}}}{1.8\times {{10}^{-5}}}
∴[NH4OH]=0.556M\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)(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−OH^- can also be found from pOHpOH where pH+pOH=14pH+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.