So \(\text{NH}_{3}\) and \(\text{NH}_{4}^{+}\) are a \({\color{purple}{\textbf{conjugate acid-base pair}}}\).The reaction between ammonia and water is another example:A conjugate acid-base pair contains two compounds that differ only by a hydrogen ion (\(\text{H}^{+}\)) and a charge of \(\text{+1}\).Determine the conjugate acid-base pairs for the following reaction:\(\text{HNO}_{3}(\text{aq}) + \text{OH}^{-}(\text{aq})\) \(\rightleftharpoons\) \(\text{NO}_{3}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)\(\text{HNO}_{3}\) is nitric acid. These are:The dissociation of bases is similar to that of acids in that we look at the \(\color{blue}{\text{K}_{\text{b}}}\) values in a similar manner:\(\color{blue}{\text{NaOH(aq)}}\) + \(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{Na}^{+}(\text{aq}) + \text{OH}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)\(\color{blue}{\text{K}_{\text{b}}} = \dfrac{\text{[Na}^{+}{\text{(aq)][OH}}^{-}{\text{(aq)]}}}{\color{blue}{\text{[NaOH(aq)]}}}\)The \(\color{blue}{\text{K}_{\text{b}}}\) for \(\text{NaOH}\) is very large (\(\text{NaOH}\) is a strong base and almost completely dissociates) and shows that the equilibrium lies very far to the \(\text{OH}^{-}\) side of the reaction. It reacts with the acids in dough to release carbon dioxide. As a result we can write the equilibrium as:\(\color{blue}{\text{NH}_{3}{\text{(g)}}}\) + \(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{NH}_{4}^{+}(\text{aq}) + \text{OH}^{-}(\text{aq})\)\(\color{blue}{\text{K}_{\text{b}}} = \dfrac{\text{[NH}_{4}^{+}{\text{(aq)][OH}}^{-}{\text{(aq)]}}}{\color{blue}{\text{[NH}_{3}{\text{(g)]}}}}\)The \(\color{blue}{\text{K}_{\text{b}}}\) for \(\text{NH}_{3}\) is approximately \(\text{1,8} \times \text{10}^{-\text{5}}\) (\(\text{NH}_{3}\) is a weak base) and shows that the equilibrium lies to the \(\text{NH}_{3}\) side of the reaction. This is a high ratio, therefore the solution of \(\text{HCl}\) is concentrated.\(\text{0,01}\) \(\text{g}\) of \(\text{Mg}(\text{OH})_{2}\) is added to \(\text{1 000}\) \(\text{dm$^{3}$}\) of water. The number of moles (n) needs to be calculated. Only a small percentage of the \(\text{HF}\) molecules ionise in the solution.\(\text{NaOH}\) is sodium hydroxide and is a base. Only a small amount of ionisation or dissociation means an acid or base is weak.Almost all the \(\text{HCl}\) molecules ionise in the solution, therefore \(\text{HCl}\) is a strong acid.Only a small percentage of the \(\text{Mg}(\text{OH})_{2}\) molecules dissociate, therefore \(\text{Mg}(\text{OH})_{2}\) is a weak base.A concentrated solution has a high ratio of solute to solvent. The Hydrogen atoms contain only one proton. A dilute solution can also be made from a strong or a weak acid or base. Oils and other non-aqueous liquids are not acids or bases. The pH of the final solution is \(\text{6,5}\).Is the acid a strong or a weak acid? The Arrhenius and Brønsted-Lowry models for acids and bases are discussed. As a result we can write the equilibrium as:Calculate the equilibrium constant for hydrochloric acid added to \(\text{1,38}\) \(\text{dm$^{3}$}\) of water:\(\text{HCl}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{H}_{3}\text{O}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq})\)n(\(\text{HCl}\)) in solution = \(\text{0,005}\) \(\text{mol}\)n(\(\text{Cl}^{-}\)) in solution = \(\text{87,3}\) \(\text{mol}\)C(\(\text{HCl}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0,005} {\text{ mol}}}{\text{1,38} {\text{ dm}}^{3}}\) = \(\text{0,0036}\) \(\text{mol.dm$^{-3}$}\)C(\(\text{Cl}^{-}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{87,3} {\text{ mol}}}{\text{1,38} {\text{ dm}}^{3}}\) = \(\text{63,3}\) \(\text{mol.dm$^{-3}$}\)There is a \(\text{1}\):\(\text{1}\) mole ratio between \(\text{H}_{3}\text{O}^{+}\) and \(\text{Cl}^{-}\).

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