# 7.13: Calculating Ka and Kb

The pH meter was invented because Florida orange growers needed a way to test the acidity of their fruit. The first meter was invented by Arnold Beckman, who went on to form Beckman Instruments. Beckman's business was very successful, and he used much of his fortune to fund science education and research. The Beckman family donated \$40 million to build the Beckman Institute at the University of Illinois.

## Calculating $$K_\text{a}$$ and $$K_\text{b}$$

The numerical value of $$K_\text{a}$$ and $$K_\text{b}$$ can be determined from an experiment. A solution of known concentration is prepared and its pH is measured with an instrument called a pH meter. Figure $$\PageIndex{1}$$: A pH meter is a laboratory device that provides quick, accurate measurements of the pH of solutions. (CC BY-NC; CK-12)

Example $$\PageIndex{1}$$

A $$0.500 \: \text{M}$$ solution of formic acid is prepared and its pH is measured to be 2.04. Determine the $$K_\text{a}$$ for formic acid.

Solution

Step 1: List the known values and plan the problem.

Known

• Initial $$\left[ \ce{HCOOH} \right] = 0.500 \: \text{M}$$
• pH $$= 2.04$$

Unknown

• $$K_\text{a} = ?$$

First, the pH is used to calculate the $$\left[ \ce{H^+} \right]$$ at equilibrium. An ICE table is set up in order to determine the concentrations of $$\ce{HCOOH}$$ and $$\ce{HCOO^-}$$ at equilibrium. All concentrations are then substituted into the $$K_\text{a}$$ expression and the $$K_\text{a}$$ value is calculated.

Step 2: Solve.

$\left[ \ce{H^+} \right] = 10^{-\text{pH}} = 10^{-2.04} = 9.12 \times 10^{-3} \: \text{M}$

Since each formic acid molecule that ionizes yields one $$\ce{H^+}$$ ion and one formate ion $$\left( \ce{HCOO^-} \right)$$, the concentrations of $$\ce{H^+}$$ and $$\ce{HCOO^-}$$ are equal at equilibrium. We assume that the initial concentrations of each ion are zero, resulting in the following ICE table.

$\begin{array}{l|ccc} & \ce{HCOOH} & \ce{H^+} & \ce{HCOO^-} \\ \hline \text{Initial} & 0.500 & 0 & 0 \\ \text{Change} & -9.12 \times 10^{-3} & +9.12 \times 10^{-3} & +9.12 \times 10^{-3} \\ \text{Equilibrium} & 0.491 & 9.12 \times 10^{-3} & 9.12 \times 10^{-3} \end{array}$

Now, substituting into the $$K_\text{a}$$ expression gives:

$K_\text{a} = \frac{\left[ \ce{H^+} \right] \left[ \ce{HCOO^-} \right]}{\left[ \ce{HCOOH} \right]} = \frac{\left( 9.12 \times 10^{-3} \right) \left( 9.12 \times 10^{-3} \right)}{0.491} = 1.7 \times 10^{-4}$

The value of $$K_\text{a}$$ is consistent with that of a weak acid. Two significant figures are appropriate for the answer, since there are two digits after the decimal point in the reported pH.

Similar steps can be taken to determine the $$K_\text{b}$$ of a base. For example, a $$0.750 \: \text{M}$$ solution of the weak base ethylamine $$\left( \ce{C_2H_5NH_2} \right)$$ has a pH of 12.31.

$\ce{C_2H_5NH_2} + \ce{H_2O} \rightleftharpoons \ce{C_2H_5NH_3^+} + \ce{OH^-}$

Since one of the products of the ionization reaction is the hydroxide ion, we need to first find the $$\left[ \ce{OH^-} \right]$$ at equilibrium. The pOH is $$14 - 12.31 = 1.69$$. The $$\left[ \ce{OH^-} \right]$$ is then found from $$10^{-1.69} = 2.04 \times 10^{-2} \: \text{M}$$. The ICE table is then set up as shown below.

$\begin{array}{l|ccc} & \ce{C_2H_5NH_2} & \ce{C_2H_5NH_3^+} & \ce{OH^-} \\ \hline \text{Initial} & 0.750 & 0 & 0 \\ \text{Change} & -2.04 \times 10^{-2} & +2.04 \times 10^{-2} & +2.04 \times 10^{-2} \\ \text{Equilibrium} & 0.730 & 2.04 \times 10^{-2} & 2.04 \times 10^{-2} \end{array}$

Substituting into the $$K_\text{b}$$ expression yields the $$K_\text{b}$$ for ethylamine.

$K_\text{b} = \frac{\left[ \ce{C_2H_5NH_3^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{C_2H_5NH_2} \right]} = \frac{\left( 2.04 \times 10^{-2} \right) \left( 2.04 \times 10^{-2} \right)}{0.730} = 5.7 \times 10^{-4}$

## Summary

• Calculations of $$K_\text{a}$$ and $$K_\text{b}$$ are described.