Question

A \(0.37 M\) solution of a weak acid (HA) has a pH of \(3.7\). What is the \(K_{\mathrm{a}}\) for this acid?

Short answer

\(K_a \approx 5.01 \times 10^{-8}\)

Step-by-step solution

- Write the dissociation equation

Write the dissociation equation for the weak acid HA: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

- Understand the relationship between pH and \(\text{H}^+\) concentration

Remember the relationship between pH and \(\text{H}^+\) concentration. Use the formula: \[ \text{pH} = -\text{log}[\text{H}^+] \]Given that the pH is 3.7, calculate \([\text{H}^+]\).

- Calculate \([\text{H}^+]\)

Rearrange the formula to find \([\text{H}^+]\): \[ [\text{H}^+] = 10^{-\text{pH}} \] Plug in the given pH value: \[ [\text{H}^+] = 10^{-3.7} \] Use a calculator to find \([\text{H}^+]\).

- Set up the equilibrium expression for \(\text{K}_a\)

For the dissociation of the weak acid: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] The expression for \(\text{K}_a\) is: \[ \text{K}_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \] Since each molecule of HA produces one \(\text{H}^+\) and one \(\text{A}^-\), we have \([\text{A}^-] = [\text{H}^+]\).

- Plug in the known values

The initial concentration of HA is given as 0.37 M. The concentration of \(\text{H}^+\) is found from the previous step. Plug these into the expression: \[ \text{K}_a = \frac{(10^{-3.7})(10^{-3.7})}{0.37 - 10^{-3.7}} \] Since \([\text{H}^+]\) is very small compared to the initial concentration of HA, \(\text{HA\textsubscript{initial}} - [\text{H}^+]) \approx 0.37\).

- Simplify the expression and calculate \(\text{K}_a\)

Simplify the expression: \[ \text{K}_a = \frac{(10^{-3.7})^2}{0.37} \] Calculate \(\text{K}_a\): \[ \text{K}_a \approx 5.01 \times 10^{-8} \]

Key concepts

Weak Acids

Weak acids only partially dissociate in water. This means that only a small fraction of the acid molecules release hydrogen ions (\text{H}^+).
The dissociation can be represented by an equilibrium equation, showing a balance between the undissociated acid and the ions produced.
For example, the dissociation of a general weak acid HA in water can be written as:
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]
This equilibrium position indicates that both the forward and reverse reactions occur at a constant rate, keeping the concentrations of all species involved steady over time.
Because weak acids do not fully dissociate, their \text{H}^+ concentrations are much lower compared to strong acids at the same concentration.
Understanding this principle is crucial for calculating the acid dissociation constant, \text{K}_a, which quantifies the degree of dissociation.

pH Calculation

The pH of a solution measures its acidity or hydrogen ion concentration. It is calculated using the following formula:
\[\text{pH} = -\text{log}[\text{H}^+]\]
This equation shows that pH is the negative logarithm of the hydrogen ion concentration. By knowing the pH, we can determine \([\text{H}^+]\) through rearrangement:
\[\text{H}^+] = 10^{-\text{pH}}\]
In the given problem, the pH is 3.7, so:
\[\text{H}^+ = 10^{-3.7} \text{M}\]
Using a calculator, we find that \([\text{H}^+]\) is about \(2.0 \times 10^{-4}\text{M}\).
This calculated \([\text{H}^+]\) helps us understand the extent of dissociation and is essential for finding the \(\text{K}_a\) of the weak acid.

Equilibrium Expression

The equilibrium expression for the acid dissociation constant, \(\text{K}_a\), reveals the relationship between the concentrations of the products and the reactants at equilibrium.
For the weak acid HA, the dissociation equilibrium is:
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]
The corresponding equilibrium expression is:
\[\text{K}_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]
Here, \([\text{H}^+]\) and \([\text{A}^-]\) represent the concentrations of hydrogen ions and the conjugate base, respectively, and \([\text{HA}]\) is the concentration of the undissociated acid.
Because one molecule of HA dissociates to form one \(\text{H}^+\) and one \(\text{A}^-\), we know that \([\text{A}^-] = [\text{H}^+]\).
By substituting the values from the given pH calculation and the original concentration of HA (0.37 M), the \(\text{K}_a\) expression simplifies to:
\(\text{K}_a = \frac{ (10^{-3.7})^2}{0.37} \text{M}\)
Thus, \(\text{K}_a\) is calculated to be approximately \(5.01 \times 10^{-8}\).
Understanding this process helps in determining the strength of the weak acid.