Chemistry » Acid-Base and Redox Reactions » Acids And Bases Continued

Acid-Base Equilibrium Constants

\(K_{a}\) and \(K_{b}\)

The equilibrium constant for the ionisation process of an acid (the extent to which ions are formed in solution) is given by the term \({\text{K}_{\color{red}{\textbf{a}}}}\), while that for a base is given by \({\text{K}_{\color{blue}{\textbf{b}}}}\). These equilibrium constants are a way of determining whether the acid or base is weak or strong.

Remember from Chemistry 109 that we calculate K as follows:

\({\text{K}} = \dfrac{\text{[products]}}{\text{[reactants]}}\)

Acids

  • Strong acids

    Consider the ionisation of \(\text{HBr}\):

    \(\color{red}{\text{HBr(g)}}\) + \(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{H}_{3}\text{O}^{+}(\text{aq}) + \text{Br}^{-}(\text{aq})\)

    We are calculating \(\color{red}{\text{K}_{\text{a}}}\) here as it is an \(\color{red}{\text{acid}}\) being dissolved in water. The liquid water is not included in the equation.

    \(\color{red}{\text{K}_{\text{a}}} = \dfrac{\text{[H}_{3}{\text{O}}^{+}{\text{(aq)][Br}^{-}{\text{(aq)]}}}}{\color{red}{\text{[HBr(g)]}}}\)

    The value of \(\text{K}_{\text{a}}\) for this reaction is very high (approximately \(\text{1} \times \text{10}^{\text{9}}\)). This means that there are many more moles of product than reactant. The \(\text{HBr}\) molecules are almost all ionised to \(\text{H}^{+}\) and \(\text{Br}^{-}\).

    We can then show that in the ionisation reaction \(\text{HBr}\) is almost completely ionised. Effectively, any value above \(\text{1} \times \text{10}^{\text{3}}\) is large and shows that almost complete ionisation has occurred:

    03e53103a394bb06535cce32b30b06fc.png

    The unequal double arrows in the reaction equation indicate that the equilibrium position favours the formation of ions. This is in contrast to the ionisation of a weak acid.

  • Weak acids

    Consider the ionisation of ethanoic acid (\(\text{CH}_{3}\text{COOH}\)):

    \(\color{red}{\text{CH}_{3}{\text{COOH(aq)}}}\) + \(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{H}_{3}\text{O}^{+}(\text{aq}) + \text{CH}_{3}\text{COO}^{-}(\text{aq})\)

    Therefore, \(\color{red}{\text{K}_{\text{a}}} = \dfrac{\text{[H}_{3}{\text{O}}^{+}{\text{(aq)][CH}}_{3}{\text{COO}}^{-}{\text{(aq)]}}}{\color{red}{\text{[CH}_{3}{\text{COOH(aq)]}}}}\)

    The value of \(\color{red}{\text{K}_{\text{a}}}\) for this reaction is very low (approximately \(\text{1.7} \times \text{10}^{-\text{5}}\)), showing that only a few of the \(\text{CH}_{3}\text{COOH}\) molecules ionise to form \(\text{H}_{3}\text{O}^{+}\) (in water) and \(\text{CH}_{3}\text{COO}^{-}\).

    We can write the reaction with an unequal double arrow to show the position of the equilibrium, which does not favour the formation of ions:

    5b8a4e2144c59ba4eb5a93d94ba9a576.png

Name

Formula

\(\text{K}_{\text{a}}\) values

Type

Hydrobromic acid

\(\text{HBr}\)

\(\text{1.0} \times \text{10}^{\text{9}}\)

strong acid

Hydrochloric acid

\(\text{HCl}\)

\(\text{1.3} \times \text{10}^{\text{6}}\)

strong acid

Sulfuric acid

\(\text{H}_{2}\text{SO}_{4}\)

First \(\text{H}^{+}\): \(\text{1.0} \times \text{10}^{\text{3}}\)

Second \(\text{H}^{+}\): \(\text{1.0} \times \text{10}^{-\text{2}}\)

strong acid

Oxalic acid

\(\text{H}_{2}\text{C}_{2}\text{O}_{4}\)

First \(\text{H}^{+}\): \(\text{5.8} \times \text{10}^{-\text{2}}\)

Second \(\text{H}^{+}\): \(\text{6.5} \times \text{10}^{-\text{5}}\)

weak acid

Sulfurous acid

\(\text{H}_{2}\text{SO}_{3}\)

First \(\text{H}^{+}\): \(\text{1.4} \times \text{10}^{-\text{2}}\)

Second \(\text{H}^{+}\): \(\text{6.3} \times \text{10}^{-\text{8}}\)

weak acid

Hydrofluoric acid

\(\text{HF}\)

\(\text{3.5} \times \text{10}^{-\text{4}}\)

weak acid

Ethanoic acid

\(\text{CH}_{3}\text{COOH}\)

\(\text{1.7} \times \text{10}^{-\text{5}}\)

weak acid

Table: \(\text{K}_{\text{a}}\) values for the ionisation of some common acids. Note that sulfuric, oxalic and sulfurous acid can all lose \(\text{2}\) \(\text{H}^{+}\) ions (they are diprotic acids).

Fact:

There are really only six strong inorganic acids, the rest are considered weak. These are:

\(\text{HClO}_{4}\) (perchloric acid),

\(\text{HI}\) (hydroiodic acid),

\(\text{HBr}\) (hydrobromic acid),

\(\text{HCl}\) (hydrochloric acid),

\(\text{H}_{2}\text{SO}_{4}\) (sulfuric acid) and

\(\text{HNO}_{3}\) (nitric acid).

Bases

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:

  • Strong bases

    For a strong base like \(\text{NaOH}\):

    \(\color{blue}{\text{NaOH(aq)}}\) + \(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{Na}^{+}(\text{aq}) + \text{OH}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)

    \(\color{blue}{\text{K}_{\text{b}}} = \dfrac{\text{[Na}^{+}{\text{(aq)][OH}}^{-}{\text{(aq)]}}}{\color{blue}{\text{[NaOH(aq)]}}}\)

    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. As a result we can write the equilibrium as:

    791619afd5ebc92c3b0b31474a1f82b3.png

  • Weak bases

    \(\color{blue}{\text{NH}_{3}{\text{(g)}}}\) + \(\text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{NH}_{4}^{+}(\text{aq}) + \text{OH}^{-}(\text{aq})\)

    \(\color{blue}{\text{K}_{\text{b}}} = \dfrac{\text{[NH}_{4}^{+}{\text{(aq)][OH}}^{-}{\text{(aq)]}}}{\color{blue}{\text{[NH}_{3}{\text{(g)]}}}}\)

    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. As a result we can write the equilibrium as:

    098d68ca79fc2a22aa069c18946a328b.png

Example: Equilibrium Constant Calculations

Question

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 = \(\text{87.3}\) \(\text{mol}\)

Step 1: Calculate the concentration of \(\text{HCl}\) at equilibrium

C(\(\text{HCl}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.005} {\text{ mol}}}{\text{1.38} {\text{ dm}}^{3}}\) = \(\text{0.0036}\) \(\text{mol.dm$^{-3}$}\)

Step 2: Calculate the concentration of \(\text{Cl}^{-}\) at equilibrium

C(\(\text{Cl}^{-}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{87.3} {\text{ mol}}}{\text{1.38} {\text{ dm}}^{3}}\) = \(\text{63.3}\) \(\text{mol.dm$^{-3}$}\)

Step 3: Calculate the concentration of \(\text{H}_{3}\text{O}^{+}\) at equilibrium

There is a \(\text{1}\):\(\text{1}\) mole ratio between \(\text{H}_{3}\text{O}^{+}\) and \(\text{Cl}^{-}\). Therefore, if \(\text{87.3}\) \(\text{mol}\) of \(\text{Cl}^{-}\) is present at equilibrium, \(\text{87.3}\) \(\text{mol}\) of \(\text{H}_{3}\text{O}^{+}\) must be present as well.

C(\(\text{H}_{3}\text{O}^{+}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{87.3} {\text{ mol}}}{\text{1.38} {\text{ dm}}^{3}}\) = \(\text{63.3}\) \(\text{mol.dm$^{-3}$}\)

Step 4: Calculate \(\text{K}_{\text{a}}\) (because \(\text{HCl}\) is an acid)

\(\text{K}_{\text{a}} = \dfrac{\text{[H}_{3}{\text{O}}^{+}{\text{(aq)]}}^{1}{\text{[Cl}}^{-}{\text{(aq)]}}^{1}}{\text{[HCl]}^{1}}\)

\(\text{K}_{\text{a}} = \dfrac{\text{63.3} \times \text{63.3}}{\text{0.0036}}\) = \(\text{1.11} \times \text{10}^{\text{6}}\)

Tip:

In a balanced chemical equation such as:

\(2\text{C}\) \(\to\) \(\text{A} + \text{B}\)

\(\text{K}_{\text{a}}\) = \(\dfrac{\text{[A]}^{1}\text{[B]}^{1}}{\text{[C]}^{2}}\)

However, we generally do not include the superscripts when the coefficients in the balanced equation are 1:

\(\text{K}_{\text{a}}\) = \(\dfrac{\text{[A][B]}}{\text{[C]}^{2}}\)

Remember that \(\text{x}^{1}\) = x.

Example: Equilibrium Constant Calculations

Question

The equilibrium constant (\(\text{K}_{\text{b}}\)) for the following reaction is \(\text{1.8} \times \text{10}^{-\text{5}}\).

\(\text{NH}_{3}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\) \(\to\) \(\text{NH}_{4}^{+}(\text{aq}) + \text{OH}^{-}(\text{aq})\)

Calculate the mass (at equilibrium) of \(\text{NH}_{3}\) molecules dissolved in \(\text{4}\) \(\text{dm$^{3}$}\) of water if there is a \(\text{0.00175}\) \(\text{mol.dm$^{-3}$}\) concentration of hydroxide ions at equilibrium.

Step 1: What is the \(\text{K}_{\text{b}}\) equation for this reaction

\(\text{K}_{\text{b}} = \dfrac{{\text{[NH}}_{4}^{+}{\text{(aq)]}}^{1}{\text{[OH}}^{-}{\text{(aq)]}}^{1}}{{\text{[NH}}_{3}{\text{(aq)]}}^{1}}\)

Step 2: Calculate the concentration of \(\text{NH}_{3}\) at equilibrium

If there is a \(\text{0.00175}\) \(\text{mol.dm$^{-3}$}\) concentration of \(\text{OH}^{-}\) ions at equilibrium there must be an equal concentration of \(\text{NH}_{4}^{+}\) ions.

\begin{align*} \text{[NH}_{3}{\text{(aq)]}} & = \frac{{\text{[NH}}_{4}^{+}{\text{(aq)]}}^{1}{\text{[OH}}^{-}{\text{(aq)]}}^{1}}{{\text{K}}_{\text{b}}} \\ & = \dfrac{{\text{(}}\text{0.00175}{\text{)}}{\text{(}}\text{0.00175}{\text{)}}}{\text{1.8} \times \text{10}^{-\text{5}}} \\ & = \text{0.17}\text{ mol.dm$^{-3}$} \end{align*}

Step 3: Calculate the number of moles of \(\text{NH}_{3}\) at equilibrium

\(\text{C (mol.dm}^{-3}{\text{)}} = \dfrac{\text{n (mol)}}{\text{V (dm}^{3}{\text{)}}}\), therefore:

n = C x V = \(\text{0.17}\) \(\text{mol.dm$^{-3}$}\) x \(\text{4}\) \(\text{dm$^{3}$}\) = \(\text{0.68}\) \(\text{mol}\)

Step 4: Calculate the mass of \(\text{NH}_{3}\) in solution at equilibrium

\(\text{n (mol)} = \dfrac{\text{m (g)}}{\text{M (g.mol}^{-1}{\text{)}}}\). Therefore m = n x M

M(\(\text{NH}_{3}\)) = \(\text{14.0}\) + (\(\text{3}\) x \(\text{1.01}\)) = \(\text{17.03}\) \(\text{g.mol$^{-1}$}\)

m = \(\text{0.68}\) \(\text{mol}\) x \(\text{17.03}\) \(\text{g.mol$^{-1}$}\) = \(\text{11.6}\) \(\text{g}\)

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