Question

An aqueous solution contains 10% ammonia by mass and has a density. Calculate hydroxyl and hydrogen ion concentration in this solution. (${{K}_{a}}$ for $N{{H}^{+}}=5.0\times {{10}^{-10}}M$ )

Answer 1

The base dissociation constant measures a base’s basicity and strength. The pH of a weak base in aqueous solution depends on the strength of the base and the concentration of the base. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases.

Complete answer:
Given ammonia mass, 10% means, 10g of ammonia in 100g of the solution.
Weight of ammonia, w = 10g
Weight of the solution, W = 100g
The density of the solution (given), d = $0.99g/c{{m}^{3}}$
Hence the volume of the solution, $V=\dfrac{W}{d}=\dfrac{100}{0.99}\times {{10}^{-3}}L$
Molarity of ammonia, ${{M}_{N{{H}_{3}}}}=\dfrac{10g/17g}{V}=5.82M$
$N{{H}_{3}}+{{H}_{2}}O\to N{{H}_{4}}OH$
The product ammonium hydroxide is a weak base, which is not dissociated completely.
Hence,

& Before\text{ }dissociation-\underset{1}{\mathop{N{{H}_{4}}OH}}\,\Leftrightarrow \underset{0}{\mathop{N{{H}_{4}}^{+}}}\,+\underset{0}{\mathop{O{{H}^{-}}}}\, \\\ & \text{After }dissociation-\underset{1-\alpha }{\mathop{N{{H}_{4}}OH}}\,\Leftrightarrow \underset{\alpha }{\mathop{N{{H}_{4}}^{+}}}\,+\underset{\alpha }{\mathop{O{{H}^{-}}}}\, \\\ \end{aligned}$$ Where, $\alpha $ = dissociation constant of the above reaction. The concentration of $$[O{{H}^{-}}]=C\alpha =\sqrt{{{K}_{O{{H}^{-}}}}XC}$$ Where c = concentration of ammonia = ${{M}_{N{{H}_{3}}}}$ $$\begin{aligned} & watercons\tan t\text{ }{{K}_{w}}=[{{K}_{N{{H}_{4}}+}}][{{K}_{O{{H}^{-}}}}] \\\ & {{K}_{O{{H}^{-}}}}=\dfrac{{{10}^{-14}}}{5\times {{10}^{-10}}}=2\times {{10}^{-5}} \\\ \end{aligned}$$ $$[O{{H}^{-}}]=\sqrt{2\times {{10}^{-5}}\times 5.82}=1.07\times {{10}^{-2}}M$$ $$[{{H}^{+}}]=\dfrac{{{10}^{-14}}}{[O{{H}^{-}}]}=0.092\times {{10}^{-12}}M$$ **Note:** The base dissociation constant ${{K}_{b}}$ provides a measure of the equilibrium position, If ${{K}_{b}}$ is large, the products of the dissociation reactions are favored. If ${{K}_{b}}$ is small, the undissociated base is favored. It provides a measure of the strength of a base, if this value is large, the base is largely dissociated so the base is strong, if the value is small, very little of the base is dissociated so the base is weak.