Question

Calculate the \(\mathrm{pH}\) of a solution that has an acetic acid concentration of 0.050 \(\mathrm{M}\) and a sodium acetate concentration of 0.075 M.

Short answer

The pH of the solution is approximately 4.92.

Step-by-step solution

Understand the Problem

We are asked to calculate the pH of a buffered solution containing acetic acid and sodium acetate. This is a typical buffer problem where both a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate) are present.

Identify the Relevant Equation

For a buffer solution, the pH can be determined using the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]where \( \mathrm{pK_a} \) is the negative logarithm of the acid dissociation constant (\( K_a \)) for acetic acid, \( [\text{A}^-] \) is the concentration of the acetate ion, and \( [\text{HA}] \) is the concentration of acetic acid.

Find the Acid Dissociation Constant

The acid dissociation constant \( K_a \) for acetic acid is approximately \( 1.8 \times 10^{-5} \). Therefore, to find the \( \mathrm{pK_a} \), we use the relation:\[ \mathrm{pK_a} = -\log(1.8 \times 10^{-5}) \approx 4.74 \].

Plug Values into the Henderson-Hasselbalch Equation

Substitute the known values into the Henderson-Hasselbalch equation:\[ \mathrm{pH} = 4.74 + \log \left( \frac{0.075}{0.050} \right) \].Calculate the ratio:\[ \frac{0.075}{0.050} = 1.5 \].

Calculate the pH

Now compute:\[ \mathrm{pH} = 4.74 + \log(1.5) = 4.74 + 0.176 \approx 4.916 \].Therefore, the pH of the solution is approximately 4.92.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is essential for calculating the \(\text{pH}\) of buffer solutions, like those containing acetic acid and sodium acetate.This equation provides a way to relate the \(\text{pH}\) of a solution to the concentrations of an acid and its conjugate base.The equation is:

• \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

In this formula:

• \(\text{pH}\) is what you are looking to calculate.
• \(\text{pK}_a\) represents the negative logarithm of the acid dissociation constant \(K_a\).
• \([\text{A}^-]\) is the concentration of the conjugate base, here acetate ion.
• \([\text{HA}]\) is the concentration of the weak acid, here acetic acid.

The beauty of this equation is its simplicity.You only need to plug in the known concentrations and calculate a simple logarithm.This makes it much easier than solving complex equilibrium equations.By understanding the relationships between these variables, you can determine how changing concentrations of either component affects the \(\text{pH}\).This is particularly useful in many biological and chemical processes where maintaining a specific \(\text{pH}\) is necessary.

Acid Dissociation Constant

The acid dissociation constant, represented as \(K_a\), is a critical concept when studying acids and their strengths.It helps in predicting the degree to which an acid can donate its protons to water, forming hydronium ions.The equation for the dissociation of a weak acid like acetic acid can be written as:\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] Where \( \text{HA} \) represents the acetic acid, \( \text{H}^+ \) is the proton or hydronium ion, and \( \text{A}^- \) is the acetate ion.The constant \(K_a\) is given by:\[ \text{K}_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]An acid with a high \(K_a\) value donates protons easily and is stronger, while a low \(K_a\) indicates a weaker acid.Acetic acid, with a \(K_a\) of about \(1.8 \times 10^{-5}\), is a weak acid.To make calculations easier, we use \(\text{pK}_a\), which is the negative log of \(K_a\):

• \[ \text{pK}_a = -\log(\text{K}_a) \]

For acetic acid, \(\text{pK}_a\) is approximately 4.74.This offers a more user-friendly number to work with in equations and is used in the Henderson-Hasselbalch equation for calculating \(\text{pH}\).

Acetic Acid and Sodium Acetate Buffer

Acetic acid and sodium acetate form a classic buffer system important in maintaining \(\text{pH}\) stability in solutions.Buffers resist drastic changes in \(\text{pH}\) when small amounts of acids or bases are added.This characteristic makes them valuable in both laboratory and biological settings.
The buffer works because acetic acid (\(\text{CH}_3\text{COOH}\)) is a weak acid that partially dissociates in water, while the acetate ion (\(\text{CH}_3\text{COO}^-\)) acts as its conjugate base.A typical example involves the addition of sodium acetate, which dissociates fully in water to provide acetate ions.The presence of both acetic acid and acetate ions establishes an equilibrium:

• \[ \text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^- \]

This equilibrium creates a buffer capacity that stabilizes the \(\text{pH}\).
When an acid is added to the buffer, acetate ions react with added \(\text{H}^+\), limiting the decrease in \(\text{pH}\).Conversely, if a base is added, the acetic acid would donate more \(\text{H}^+\) ions, lessening the \(\text{pH}\) change.Using the Henderson-Hasselbalch equation, we calculate the \(\text{pH}\) of this buffer by plugging in the concentrations of acetic acid and acetate ions.