Question

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.10 M\) acetic acid \(/ 0.25 M\) sodium acetate b. \(0.25 M\) acetic acid/0.10 \(M\) sodium acetate c. \(0.080 M\) acetic acid \(/ 0.20 M\) sodium acetate d. \(0.20 M\) acetic acid/0.080 \(M\) sodium acetate

Short answer

The pH values of the buffered solutions are: a. \(pH \approx 5.14\) b. \(pH \approx 4.34\) c. \(pH \approx 5.44\) d. \(pH \approx 4.04\)

Step-by-step solution

Identify the concentration of each component in the buffered solution

For each buffered solution, we are given the respective molar concentrations of acetic acid (HA) and sodium acetate (A-). We will use these concentrations to plug into the Henderson-Hasselbalch equation. a. \(0.10 M\) acetic acid (HA) / \(0.25 M\) sodium acetate (A-) b. \(0.25 M\) acetic acid (HA) / \(0.10 M\) sodium acetate (A-) c. \(0.080 M\) acetic acid (HA) / \(0.20 M\) sodium acetate (A-) d. \(0.20 M\) acetic acid (HA) / \(0.080 M\) sodium acetate (A-)

Apply the Henderson-Hasselbalch equation

Using the given pKa value for acetic acid (4.74) and the respective concentrations for each buffered solution, calculate the pH using the Henderson-Hasselbalch equation: \[pH = pK_a + \log \frac{[A^-]}{[HA]}\] a. pH = 4.74 + \(\log{\frac{0.25}{0.10}}\) b. pH = 4.74 + \(\log{\frac{0.10}{0.25}}\) c. pH = 4.74 + \(\log{\frac{0.20}{0.080}}\) d. pH = 4.74 + \(\log{\frac{0.080}{0.20}}\)

Calculate the pH of each buffered solution

Calculate the values for each of the pH expressions obtained in step 2. a. pH = 4.74 + \(\log{(2.5)}\) ≈ 4.74 + 0.4 = 5.14 b. pH = 4.74 + \(\log{(0.4)}\) ≈ 4.74 - 0.4 = 4.34 c. pH = 4.74 + \(\log{(2.5)}\) ≈ 4.74 + 0.7 = 5.44 d. pH = 4.74 + \(\log{(0.4)}\) ≈ 4.74 - 0.7 = 4.04

Summarize the results

The calculated pH values for the buffered solutions are as follows: a. \(pH \approx 5.14\) b. \(pH \approx 4.34\) c. \(pH \approx 5.44\) d. \(pH \approx 4.04\)

Key concepts

Buffered Solutions pH Calculation

Understanding the pH of buffered solutions is crucial for many scientific applications. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. This property is due to the presence of a weak acid and its conjugate base or a weak base and its conjugate acid. To calculate the pH, we commonly use the Henderson-Hasselbalch equation:

\[\begin{equation}pH = pK_a + \text{log}\frac{[A^{-}]}{[HA]}\end{equation}\]
where \(pK_a\) is the acid dissociation constant, \([A^{-}]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid.

• Identify the \(pK_a\) value of the weak acid in the buffer.
• Determine the concentrations of the acid \([HA]\) and its conjugate base \([A^{-}]\).
• Plug these values into the Henderson-Hasselbalch equation to find the pH.

By using these steps, students can gain a clear understanding of the process involved in calculating the pH of buffered solutions.

Acetic Acid and Sodium Acetate Buffer

A classic example of a buffered solution is a mixture of acetic acid and sodium acetate. Acetic acid \(CH_3COOH\), a weak acid, partially dissociates in water, creating the acetate ion \(CH_3COO^{-}\), which is its conjugate base. Sodium acetate \(CH_3COONa\), on the other hand, is a salt that dissociates completely in water to give the same acetate ion \(CH_3COO^{-}\), and sodium ions \(Na^{+}\), which do not affect the pH.

The combination of these two substances creates a buffer because the acetate ions can react with any added hydrogen ions \(H^{+}\), preventing significant changes in pH, while additional hydroxide ions \(OH^{-}\) can be neutralized by the acetic acid.

• The acid component \( HA \) in the equation represents acetic acid.
• The base component \( A^{-} \) represents the acetate ions provided by sodium acetate.

This relationship allows us to calculate the pH of the buffer solution using the concentrations of acetic acid and sodium acetate.

pKa Value

The \(pK_a\) value is an essential term in acid-base chemistry, representing the negative logarithm of the acid dissociation constant \(K_a\). It measures the strength of an acid in solution: the lower the \(pK_a\) value, the stronger the acid. For acetic acid, the \(pK_a\) is 4.74, which indicates it is a relatively weak acid since it's much higher than 0.

The \(pK_a\) value comes in handy when using the Henderson-Hasselbalch equation, as it allows for an easy-to-perform logarithmic calculation to determine the pH of buffered solutions:

• Buffers typically work best when their pH is close to the \(pK_a\) value of the acid.
• Knowing the \(pK_a\) helps predict how the buffer will respond to the addition of acids or bases.

A strong understanding of \(pK_a\) values can greatly enhance a student's ability to predict and manipulate buffered solution behavior.

Logarithmic pH Calculation

Calculating pH involves understanding the logarithmic scale, which can be intimidating at first. The pH scale is logarithmic, meaning each whole number on the scale corresponds to a tenfold change in hydrogen ion concentration. As such, a small change in pH can indicate a significant change in acidity or basicity.

To perform a logarithmic calculation for pH, one must use the base-10 logarithm of the ratio of the concentrations of the conjugate base to the weak acid. This is a core component of the Henderson-Hasselbalch equation:

\[\begin{equation}pH = pK_a + \text{log}\frac{[A^{-}]}{[HA]}\end{equation}\]
For example, if the log ratio is positive, the pH will be greater than the \(pK_a\), indicating a basic solution. Conversely, if the log ratio is negative, the solution is acidic. Understanding logarithmic relationships is vital to accurately calculate and understand pH values in buffer systems, and mastering this can aid in numerous scientific and industrial applications involving acidity and pH control.