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: 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: 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: • 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: ## 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 $$\text{17.03}$$ $$\text{g.mol^{-1}}$$ = $$\text{11.6}$$ $$\text{g}$$