Question

Calculation of the \(\mathrm{pH}\) of a Mixture of a Weak Acid and Its Conjugate Base Calculate the \(\mathrm{pH}\) of a dilute solution that contains a molar ratio of potassium acetate to acetic acid \(\left(\mathrm{p} K_{\mathrm{a}}=4.76\right)\) of a. \(2: 1\) b. \(1: 3 ;\) c. \(5: 1\) d. \(1: 1 ;\) e. \(1: 10\).

Short answer

a. 5.06, b. 4.28, c. 5.46, d. 4.76, e. 3.76.

Step-by-step solution

Understand the Buffer Solution

The mixture of a weak acid and its conjugate base is known as a buffer solution. The pH of buffer solutions can be calculated using the Henderson-Hasselbalch equation.For a weak acid (HA) and its conjugate base (A⁻), the Henderson-Hasselbalch equation is given by:\[\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]

Specify the pK(a) and Molar Ratios

The problem provides the pKa of acetic acid, which is 4.76. For each scenario, the molar ratios of potassium acetate (which provides the conjugate base A⁻) to acetic acid are given as follows: a. 2:1 b. 1:3 c. 5:1 d. 1:1 e. 1:10

Calculate pH for Each Case Using the Henderson-Hasselbalch Equation

Substitute the given molar ratios and pKa value into the Henderson-Hasselbalch equation to find each pH value:(a) For a ratio of 2:1:\[\text{pH} = 4.76 + \log_{10}\left(\frac{2}{1}\right) = 4.76 + 0.3010 \approx 5.06\](b) For a ratio of 1:3:\[\text{pH} = 4.76 + \log_{10}\left(\frac{1}{3}\right) = 4.76 + (-0.4771) \approx 4.28\](c) For a ratio of 5:1:\[\text{pH} = 4.76 + \log_{10}\left(\frac{5}{1}\right) = 4.76 + 0.6990 \approx 5.46\](d) For a ratio of 1:1:\[\text{pH} = 4.76 + \log_{10}\left(\frac{1}{1}\right) = 4.76 + 0 = 4.76\](e) For a ratio of 1:10:\[\text{pH} = 4.76 + \log_{10}\left(\frac{1}{10}\right) = 4.76 + (-1) \approx 3.76\]

Key concepts

Buffer Solution

A buffer solution is a special type of chemical solution that resists changes in its pH level when small amounts of an acid or a base are added. It is composed of a combination of a weak acid and its conjugate base. This unique mixture ensures that the solution can "buffer" or withstand fluctuations in pH.
This happens because of the weak acid (HA) contributing protons (H⁺), while its conjugate base (A⁻) mops up any excess H⁺ ions added to the system.

• Buffer solutions are used in various scientific experiments.
• They help maintain a constant pH, which is crucial for enzyme reactions and other biochemical processes.

Because of their ability to maintain stability, buffer solutions are exploited in numerous environments, including biological systems and industrial applications.

Weak Acid

A weak acid is a type of acid that only partially dissociates in an aqueous solution. This means it does not completely separate into its ions when dissolved in water. Acetic acid (CH₃COOH) is a common example, and it is used in this exercise.

• Weak acids do not release all their hydrogen ions.
• This partial dissociation results in a relatively higher pH compared to a strong acid.

Because weak acids only partially release their hydrogen ions, they are ideal components of buffer systems. They work in tandem with their conjugate base to keep the pH from changing dramatically.

Conjugate Base

The conjugate base is what remains after a weak acid loses a hydrogen ion. In the context of buffer solutions with acetic acid, the conjugate base is acetate ( CH₃COO⁻). This base is essential because it acts to neutralize any excess hydrogen ions in the solution.

• Conjugate bases stabilize the pH by both mopping up extra protons and replenishing protons.
• The strength of the conjugate base is related to its corresponding weak acid.

The equilibrium between the weak acid and its conjugate base in a buffer solution is crucial for maintaining consistent pH levels, as it allows the solution to adjust by either donating or accepting ions.

pKa

The pKa is a numeral that represents how strong an acid is in terms of its dissociation in water. In simpler terms, it's a way to describe the tendency of an acid to give up its proton. It is the negative logarithm of the acid dissociation constant (Ka), and a smaller pKa value indicates a stronger acid.

• The Henderson-Hasselbalch equation effectively uses pKa to correlate the concept of pH with the concentration ratio of the acid and base.
• The pKa is fundamentally significant in predicting the effective pH range where a buffer solution is most efficient.

In this exercise, the pKa value of 4.76 for acetic acid helps us calculate the pH when paired with its conjugate base acetate. It provides a benchmark to determine how different ratios of acid to base will affect the pH of the buffer solution.

pH Calculation

pH calculation in buffer solutions involves the use of the Henderson-Hasselbalch equation. This equation simplifies the process of finding the pH by using the known pKa of the weak acid and the ratio of the concentrations of the acid and its conjugate base. The equation is:\[\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]
Here are some important points:

• The "log" part of the equation allows for the determination of pH over different ratios quickly.
• Using the equation involves substituting the given molar ratio and the pKa to find the solution's pH.

By applying this equation in the given exercise, we're able to determine how the different ratios of acetic acid to acetate affect the pH, helping us understand buffer effectiveness and range.