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What is the $[N{{H}_{4}}^{+}]$ in a solution containing 0.02 M $N{{H}_{3}}$(${{K}_{b}}$ = 1.8 x ${{10}^{-5}}$) and 0.01 M KOH?
(A) 1.8 x ${{10}^{-5}}$
(B) 9 x ${{10}^{-6}}$
(C) 3.6 x ${{10}^{-5}}$
(D) NONE OF THE ABOVE

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Last updated date: 13th Jun 2024
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Answer
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Hint: $N{{H}_{4}}^{+}$ ions are generated by the reaction between ammonia and hydroxide ions. The equilibrium constant for this reaction is given in the question. Write the equilibrium constant in terms of concentration of reactants and products and equate it to the value provided. With this you will arrive at the concentration of $N{{H}_{4}}^{+}$ ions.

Complete Solution :
We will write the dissociation reaction of KOH and the reaction between ammonia and hydroxide ions. This is because these are the two main reactions involved in determining the ammonium ions concentration.
Dissociation of potassium hydroxide:
$KOH\,\to \,\,{{K}^{+}}\,+\,O{{H}^{-}}$

Reaction of ammonia with hydroxide ions:
$N{{H}_{4}}OH\,\to \,\,N{{H}_{4}}^{+}\,+\,O{{H}^{-}}$

- The equilibrium constant for the above reaction is given as 1.8 x ${{10}^{-5}}$.
The concentration of hydroxide ions will be same as the original concentration of potassium hydroxide i.e. 0.01 M.
- We will now write the equilibrium concentration in terms of concentration of products and reactants and equate it to the value provided to us.
${{K}_{b}}\,=\,\dfrac{[N{{H}_{4}}^{+}]{{[O{{H}^{-}}]}^{2}}}{[N{{H}_{4}}OH]}$
${{K}_{b}}\,=\,\dfrac{[N{{H}_{4}}^{+}]{{[{{10}^{-1}}]}^{2}}}{[2\,\text{x }{{10}^{-2}}]}$
$[N{{H}_{4}}^{+}]$ = 3.6 x ${{10}^{-5}}$.
Thus, upon calculation we find that the concentration of ammonium ions in the aqueous solution is 3.6 x ${{10}^{-5}}$.
So, the correct answer is “Option C”.

Note: It is important to consider the stoichiometric coefficient of products and reactants while substituting the values in the formula for equilibrium constant. This is because the concentration is raised to the power of stoichiometric coefficient before substituting it in the formula.