Question

What is the original molarity of a solution of a weak acid whose \(K_{\mathrm{a}}\) is \(3.5 \times 10^{-5}\) and whose \(\mathrm{pH}\) is 5.26 at \(25^{\circ} \mathrm{C} ?\)

Short answer

The original molarity of the weak acid is approximately \(8.6 \times 10^{-7}\, \text{M}\).

Step-by-step solution

Understand the Problem

The problem requires us to find the original molarity (concentration) of a weak acid solution, given its acid dissociation constant \(K_a\) and \(\text{pH}\).

Convert pH to [H+]

The concentration of hydrogen ions \([\text{H}^+]\) can be found using the formula \([\text{H}^+] = 10^{-\text{pH}}\). For a \(\text{pH}\) of 5.26, \([\text{H}^+] = 10^{-5.26}\).

Calculate [H+]

Calculate \([\text{H}^+] = 10^{-5.26} = 5.5 \times 10^{-6}\).

Use ICE Table

Set up an ICE (Initial, Change, Equilibrium) table to find the relationship between \([\text{H}^+]\), the dissociation of the weak acid \(HA\), and the equilibrium concentrations. Since \([\text{H}^+]\) is given, \( [\text{A}^-] = 5.5 \times 10^{-6} \).

Apply Equilibrium Expression

The equilibrium expression for the dissociation is \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\). Substitute \([\text{H}^+]\) and \([\text{A}^-]\) into this equation, assuming \([\text{HA}] = c - [\text{H}^+]\), where \(c\) is the initial concentration.

Simplify and Solve for [HA]

Since \(\text{pH}\) value is much larger than the \(K_a\), assume \([\text{HA}] \approx c\). So, \(K_a = \frac{(5.5 \times 10^{-6})^2}{c}\). Rearranging, \(c = \frac{(5.5 \times 10^{-6})^2}{3.5 \times 10^{-5}}\).

Calculate Initial Molarity c

Evaluate \(c\) from \(c = \frac{(5.5 \times 10^{-6})^2}{3.5 \times 10^{-5}}\), which gives \(c \approx 8.6 \times 10^{-7}\, \text{M}\).

Key concepts

Weak Acid Solution

A weak acid solution is one in which the acid does not completely dissociate in water. This means that in such a solution, only a small fraction of the acid molecules release hydrogen ions (\(\text{H}^+\)). Therefore, the concentration of hydrogen ions is lower compared to strong acids that completely dissociate.
Understanding the nature of weak acids is crucial. They have a specific dissociation constant, known as the acid dissociation constant \(K_a \), which indicates the extent to which an acid can release hydrogen ions into the solution.
Key points about weak acids include:

• Partial ionization in water.
• Presence of an equilibrium position between dissociated ions and undissociated molecules.
• \(K_a \) value helps predict the strength of the acid; smaller \(K_a \) values suggest that the acid is weaker.

So, when dealing with such solutions, recognizing that not all acid molecules dissociate is essential in calculations, such as finding molarity or \(pH \).

pH Calculation

Calculating the pH of a solution is a fundamental skill in chemistry. The \(pH \) scale is logarithmic and measures the concentration of hydrogen ions in a solution. It is represented by the formula \(\text{pH} = -\log([\text{H}^+])\).
For instance, in our exercise, we were given the \(pH \) as 5.26. By substituting this into the formula \([ ext{H}^+] = 10^{-\text{pH}}\), you can calculate the concentration of hydrogen ions:

• Given \(pH = 5.26 \), the concentration is \([ ext{H}^+] = 10^{-5.26} \approx 5.5 \times 10^{-6}\).

Knowing how to convert between \(pH\) and hydrogen ion concentration is essential for evaluating the characteristics of weak acid solutions and performing further calculations, like determining the molarity of an acid.

Molarity of a Solution

Molarity, often represented as \(M\) , is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution (mol/L).
In the context of calculating the molarity of a weak acid, knowing the equilibrium concentrations of ions in the solution is vital. Let's say you need to find the original molarity (denoted as \(c \)) of a weak acid:

• First, calculate \([ ext{H}^+] \) using given \(pH\).
• Use the acid dissociation constant \(K_a \) to relate the concentrations of \([ ext{H}^+] \), \([ ext{A}^-] \), and \([ ext{HA}]\).
• Assume \([ ext{HA}] \approx c\), simplify and solve \(c = \frac{(5.5 \times 10^{-6})^2}{3.5 \times 10^{-5}} \approx 8.6 \times 10^{-7}\) M.

Understanding molarity and how to calculate it is key, not just for academics, but for practical applications in the lab and industry as well.