Problem 10 - Clinical Chemistry

Question

The \(\mathrm{pK}_{\mathrm{a}}\) for acetic acid is \(4.76 .\) If the concentration of salt is \(2 \mathrm{mmol} / \mathrm{L}\) and that of acetic acid is \(6 \mathrm{mmol} / \mathrm{L}\), what is the expected \(\mathrm{pH}\) ? a. \(4.43\) b. \(6.19\) c. 104 d. 56

Step-by-Step Solution

Answer

The expected pH is approximately 4.28.

Step 1 - Write the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to estimate the pH of a buffer solution. The equation is: \[ \text{pH} = \text{pK}_\text{a} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] where \(\text{pK}_\text{a}\) is the acid dissociation constant, \([\text{A}^-]\) is the concentration of the salt (conjugate base), and \([\text{HA}]\) is the concentration of the acid.

Step 2 - Identify Given Values

From the problem, the given values are: \( \text{pK}_\text{a} = 4.76 \)\( [\text{A}^-] = 2 \text{ mmol/L} \)\( [\text{HA}] = 6 \text{ mmol/L} \)

Step 3 - Substitute the Values into Henderson-Hasselbalch Equation

Using the values from Step 2, substitute them into the equation to find the pH: \[ \text{pH} = 4.76 + \text{log} \frac{2}{6} \]

Step 4 - Simplify the Logarithmic Expression

First, calculate the ratio inside the logarithm: \[ \frac{2}{6} = \frac{1}{3} \]Then, find the logarithm value: \[ \text{log} \frac{1}{3} = \text{log}(\frac{1}{3}) \ \text{log}(\frac{1}{3}) = \text{log}(1) - \text{log}(3) = 0 - 0.477 = -0.477 \]

Step 5 - Calculate the pH

Substitute the logarithm value back into the equation to find the pH: \[ \text{pH} = 4.76 + (-0.477) = 4.76 - 0.477 = 4.283 \] Thus, the expected pH is approximately 4.28.

Key Concepts

acid dissociation constant

The acid dissociation constant, often represented as \text{k}_\text{a}, is a measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation of an acid into its conjugate base and a hydrogen ion in an aqueous solution. The equation for this dissociation can be written as: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
Where \text{HA} represents the acid, \text{H}^+ is the hydrogen ion, and \text{A}^- is the conjugate base.
The value of \text{k}_\text{a} helps to predict the degree of ionization of the acid at a given concentration.

• A large \text{k}_\text{a} value indicates a strong acid that fully dissociates in water.
• A small \text{k}_\text{a} value indicates a weak acid that only partially dissociates.

The \text{p}K_\text{a} value, which is the negative logarithm of the \text{k}_\text{a}, provides a more practical way to express the strength of an acid: \text{pK}_\text{a} = -\text{log} \text{k}_\text{a}. This value makes it easier to compare acid strengths and is typically more convenient for calculations involving pH.

buffer solution

A buffer solution is a special type of solution that resists significant changes in pH when small amounts of an acid or a base are added. Buffers are essential in many biological and chemical processes where maintaining a stable pH is crucial.
A typical buffer solution consists of a weak acid and its conjugate base.

• For example, a mixture of acetic acid \text{(HA)} and its conjugate base acetate \text{(A}^-\text{)}.

Buffers work because the weak acid can neutralize added bases, and the conjugate base can neutralize added acids.
The Henderson-Hasselbalch equation is commonly used to describe the pH of buffer solutions: \[ \text{pH} = \text{p}K_\text{a} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] Here, the ratio \text{[\text{A}^-]/[\text{HA}]} represents the proportion of the conjugate base to the weak acid.
Understanding buffer solutions is important for predicting and controlling pH in various scientific and industrial applications.

pH calculation

pH is a scale used to specify the acidity or basicity of an aqueous solution. It is defined as the negative logarithm of the hydrogen ion concentration: \[ \text{pH} = -\text{log} [\text{H}^+] \] For buffer solutions, the Henderson-Hasselbalch equation simplifies pH calculations by relating the pH to the \text{p}K_\text{a} and the ratio of the concentrations of the conjugate base and the acid.
To calculate pH using the Henderson-Hasselbalch equation, follow these steps:

• Identify the \text{p}K_\text{a} of the weak acid.
• Determine the concentrations of the conjugate base \text{(A}^-\text{)} and the acidic form \text{(HA)}.
• Substitute these values into the equation: \[ \text{pH} = \text{p}K_\text{a} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]

In the given exercise, we used the following values: \text{p}K_\text{a} of acetic acid is 4.76, [\text{A}^-] = 2 mmol/L, and [\text{HA}] = 6 mmol/L.
Substituting these into the equation: \[ \text{pH} = 4.76 + \text{log} \frac{2}{6} = 4.76 + \text{log} (\frac{1}{3}) = 4.76 - 0.477 = 4.283 \] Hence, the pH is approximately 4.28. This illustrates how the Henderson-Hasselbalch equation is a powerful tool for estimating pH in buffer solutions.