Question

For solutions of a weak acid, a graph of \(\mathrm{pH}\) versus the logarithm of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

Short answer

The magnitude of the slope of the line for a graph of pH versus the logarithm of the initial acid concentration of a weak acid is given by \( |m| = \frac{1}{2} \log K_a \) where K_a is the acid dissociation constant.

Step-by-step solution

Write the equilibrium expression for the weak acid dissociation

For a generic weak acid (HA), its equilibrium dissociation in water can be represented as follows: \( HA + H_2 O \rightleftharpoons H_3O^{+} + A^{-} \) At equilibrium, we can define the acid dissociation constant (Ka) as the ratio of the concentration of the products to the concentration of the reactants (excluding water): \( K_a = \frac{[H_3O^+ ][A^- ]}{[HA]} \)

Define the variables for a weak acid dissociation

Let the initial concentration of the weak acid be represented by "c" moles/L, and let the change in concentration due to dissociation be "x" moles/L. At equilibrium, the concentrations of different species are as follows: - HA: \( c-x \) moles/L - \( H_3O^+ \): \( x \) moles/L - \( A^- \): \( x \) moles/L Now, substitute these concentration values into the Ka expression: \( K_a = \frac{x^2}{c-x} \)

Use the definition of pH and relationship with initial concentration

The pH of a solution is defined as the negative logarithm of the H⁺(or \( H_3O^+ \))concentration: \( pH = -\log [H_3O^+] \) Since \( [H_3O^+] = x \), we can rewrite the pH equation as: \( pH = -\log x \) The logarithm of the initial concentration is given by \( \log c \). The problem states that there is a linear relationship between pH and \( \log c \), which means that we can represent it as the following equation: \( pH = m\log c + b \) where m is the slope of the line and b is the y-intercept.

Rearrange equations to express dissociation in terms of initial concentration

First, solve the pH equation for "x": \( x = 10^{-pH} \) Now, substitute this expression for "x" in the Ka expression: \( K_a = \frac{(10^{-pH})^2}{c-10^{-pH}} \) Next, solve this last equation for pH: \( pH = -\log \sqrt{\frac{K_a(c-10^{-pH})}{c}} \)

Calculate the slope of the line

Now that we have the relationship between pH and initial concentration, we can identify the slope of the line (m) by comparing it to the equation mentioned in step 3. The slope of the line is given by the coefficient of \( \log c \), which in this case is: \(m = -\log \sqrt{\frac{K_a}{c}} = -\frac{1}{2} \log K_a\) Thus, the magnitude of the slope of the line is: \( |m| = \frac{1}{2} \log K_a \)