Problem 18 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing \(0.050 \mathrm{M}\) sodium lactate.

Step-by-Step Solution

Answer

The percent ionization of 0.250 M lactic acid is approximately \(1.18\%\). In a solution containing 0.050 M sodium lactate, the percent ionization of 0.250 M lactic acid is approximately \(0.28\%\).

Step 1: Calculate the percent ionization of 0.250 M lactic acid

We can represent the ionization of lactic acid (HA) as follows: \(HA \rightleftharpoons H^+ + A^-\) Given the initial concentration of lactic acid is 0.250 M and Ka = \(1.4 \times 10^{-4}\). Construct an ICE table for the equilibrium concentrations: Initial 0.250 M 0 0 Change -x +x +x Equilibrium 0.250-x x x Now, use the Ka expression to solve for x: \[K_a =\frac {[H^+][A^-]}{[HA]} = \frac {x \times x}{0.250-x}\] Solve the equation for x: \(1.4 \times 10^{-4} = \frac {x^2}{0.250-x}\) \(x\) is the concentration of H⁺, as the solution mostly dissociates to a limited extent, we can assume that \(x << 0.250\) Simplify and solve for x: \(1.4 \times 10^{-4} ≈ \frac {x^2}{0.250}\) \(x ≈ \sqrt{0.250 \times 1.4 \times 10^{-4}}\) \(x ≈ 0.00295\,\mathrm{M}\) Calculate the percent ionization: \(\% \textrm{ ionization } = \frac{[H^+]}{0.250\,\mathrm{M}} \times 100\) \(\% \textrm{ ionization } ≈ \frac{0.00295\,\mathrm{M}}{0.250\,\mathrm{M}} \times 100\) \(\% \textrm{ ionization } ≈ 1.18\%\)

Step 2: Calculate the percent ionization of 0.250 M lactic acid in a solution containing 0.050 M sodium lactate

Now we have a buffer solution containing both the weak acid and its salt. In this case, the initial condition changes: Initial 0.250 M 0.050 M 0.050 M Change -y y y Equilibrium 0.250-y 0.050+y 0.050+y Substitute the equilibrium concentrations into the Ka expression: \(1.4 \times 10^{-4} = \frac{((0.050+y)(y))}{(0.250-y)}\) Assuming the change is small, neglect y compared to 0.050 and 0.250: \(1.4 \times 10^{-4} ≈ \frac{0.050 \times y}{0.250}\) Solve for y: \(y ≈ \frac{1.4 \times 10^{-4} \times 0.250}{0.050}\) \(y ≈ 0.0007\,\mathrm{M}\) Calculate the percent ionization: \(\%\, \textrm{ionization} = \frac{[H^+]}{0.250\,\mathrm{M}} \times 100\) \(\%\, \textrm{ionization} ≈ \frac{0.0007\,\mathrm{M}}{0.250\,\mathrm{M}} \times 100\) \(\%\, \textrm{ionization} ≈ 0.28\%\)

Key Concepts

Understanding Lactic Acid

Lactic acid is an organic compound commonly found in muscles during intense exercise and in certain fermented foods like yogurt. It has the chemical formula \(C_3H_6O_3\) and is known as a carboxylic acid with a hydroxyl group adjacent to its carbonyl group. This makes it a hydroxy acid.
Lactic acid is a weak acid, which means it only partially dissociates in water. This partial dissociation is important in various biological functions and plays a key role in metabolism. Because it does not fully ionize in solution, lactic acid’s strength as an acid is represented by its acid dissociation constant, \(K_a\).
In chemical terms, lactic acid's ionization can be expressed by its equilibrium reaction with water:

• \(HA \rightleftharpoons H^+ + A^-\)

Here, \(HA\) represents lactic acid, \(H^+\) is the hydrogen ion, and \(A^-\) is the lactate ion when lactic acid releases a proton.

Acid Dissociation Constant (Ka)

The acid dissociation constant, known as \(K_a\), is a crucial measure in chemistry. It indicates the strength of a weak acid by showing how well the acid ionizes in water. For lactic acid, \(K_a = 1.4 \times 10^{-4}\), which is small and indicates a weak acid.
The value of \(K_a\) is derived from the equilibrium expression of an acid's dissociation:

• \(K_a = \frac{[H^+][A^-]}{[HA]}\)

Here, \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of the conjugate base (lactate ion for lactic acid), and \([HA]\) is the concentration of the undissociated acid.
In the context of the exercise, knowing the \(K_a\) allows calculations of how much an acid will ionize at given conditions. This forms the basis for understanding and calculating percent ionization in solutions with and without buffers.

Buffer Solution and Its Role

A buffer solution is a mix of a weak acid and its conjugate base, or a weak base and its conjugate acid. It helps resist changes in pH when small amounts of acids or bases are added. In our exercise, a buffer includes lactic acid and its conjugate base, sodium lactate.
The ability of a buffer to stabilize pH is due to the equilibrium between the weak acid and its base. When an acid is added to the solution, the base part of the buffer neutralizes it. Conversely, when a base is added, the acid part of the buffer neutralizes that. This equilibrium helps minimize changes in hydrogen ion concentration, maintaining a steady pH.
Using the Henderson-Hasselbalch equation can simplify pH calculations in buffer solutions:

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

This equation links the pH to the \(K_a\) through the \(pK_a\) (which is \(-\log K_a\)) and describes how the concentration ratios of acid and base dictate the pH in the buffer solution.