Problem 51 - Chemistry: The Central Science in SI Units, Global Edition

Question

Lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right)\) has one acidic hydrogen. A \(0.10 \mathrm{M}\) solution of lactic acid has a pH of 2.44. Calculate \(K_{a}\)

Step-by-Step Solution

Answer

The \(K_{a}\) value for lactic acid in the given solution is approximately \(1.3 \times 10^{-4}\).

Step 1: Calculate the H+ ion concentration

We can use the formula pH = \(-\log_{10}([\mathrm{H}^+])\) to calculate the concentration of \(\mathrm{H}^+\) ions in the solution. Rearranging the formula for calculating \([\mathrm{H}^+]\): \([\mathrm{H}^+] = 10^{-\text{pH}}\) Now, let's plug in pH value given in the problem into the above formula: \([\mathrm{H}^+] = 10^{-2.44} \approx 3.6 \times 10^{-3} \, \mathrm{M}\)

Step 2: Write the equilibrium expression for lactic acid

Next, let's write out the equilibrium of lactic acid dissociation: \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COO}^{-} + \mathrm{H}^{+}\) As given, the initial concentration of lactic acid is \(0.10 \; \mathrm{M}\). Let x represent the change in concentrations at equilibrium: [CH3CH(OH)COOH] = 0.10 - x [CH3CH(OH)COO-] = x [H+] = x + 3.6 x 10^{-3} Since we know the concentration of H+ from Step 1, we can find the concentrations of the dissociated components at equilibrium.

Step 3: Express Ka in terms of x and given concentrations

The equilibrium expression is given by the formula for \(K_{a}\): \(K_{a} = \dfrac{[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COO}^{-}][\mathrm{H}^+]}{[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COOH}]}\) Now we can substitute the values that we found in Step 2 into the equilibrium expression: \(K_{a} = \dfrac{(x)(x + 3.6 \times 10^{-3})}{0.10 - x}\)

Step 4: Approximations and solving for Ka

Since the pH is 2.44, the dissociation of lactic acid is very small, and thus we can assume that x is much smaller than the initial concentration, i.e., \(x << 0.10\). We can further simplify the expression and solve for \(K_{a}\). \(K_{a} = \dfrac{(x)(x + 3.6 \times 10^{-3})}{0.10}\) Now, we know the H+ concentration from Step 1, which is approximately 3.6 x 10^{-3}: \(K_{a} = \dfrac{(3.6 \times 10^{-3})(3.6 \times 10^{-3})}{0.10} \) Finally, let's calculate the K\(_{a}\) value: \(K_{a} \approx 1.3 \times 10^{-4}\) Therefore, the \(K_{a}\) value for lactic acid in the given solution is approximately \(1.3 \times 10^{-4}\).

Key Concepts

Lactic Acid

Lactic acid is an organic compound that plays a vital role in various biochemical processes. It is often found in muscles during intense exercise, but also used in the food and cosmetic industries. The chemical formula for lactic acid is \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COOH} \). This acid is characterized by having only one acidic hydrogen atom, which can dissociate in water to produce hydrogen ions (H\(^+\)).

Being a weak acid, lactic acid does not completely dissociate in solution. Instead, it establishes an equilibrium between the undissociated acid and its corresponding ions, which include the lactate ion (\( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COO}^{-} \)) and hydrogen ion (H\(^+\)). Understanding this equilibrium helps us to calculate the acid dissociation constant, \( K_{a} \).

This ability to partially dissociate influences its behavior in solutions and consequently, its pH. In simple terms, experimenting with lactic acid in solutions offers excellent hands-on learning opportunities regarding acids and bases.

pH Calculation

Calculating the pH of a solution is an important skill in chemistry, and it provides insights into the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration, given by the formula: \( \text{pH} = -\log_{10}([\mathrm{H}^+]) \).

In this exercise, we're provided with the pH value of 2.44 for a 0.10 M solution of lactic acid. Our job is to connect the pH to the hydrogen ion concentration in the solution. To do this, we rearrange the pH formula to find \([\mathrm{H}^+]\):

• \([\mathrm{H}^+] = 10^{-2.44} \approx 3.6 \times 10^{-3} \, \mathrm{M}\)

This calculation is crucial because it allows you to determine the acidity of the solution in its simplest form, paving the way for further calculations such as those involving the \( K_{a} \) value.

Equilibrium Expression

Once the \( [\mathrm{H}^+] \) is known, the next step is to set up the equilibrium expression for the dissociation of lactic acid. When lactic acid dissolves in water, an equilibrium is established between the undissociated acid and its ions:

\[ \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COO}^{-} + \mathrm{H}^{+} \]

The initial concentration of lactic acid is given as 0.10 M, and at equilibrium, we denote the changes in concentrations with \( x \). It's important to express the concentrations in terms of \( x \):

• \([\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COOH}] = 0.10 - x\)
• \([\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COO}^{-}] = x\)
• \([\mathrm{H}^+] = x + 3.6 \times 10^{-3}\)

The equilibrium expression takes the form of the acid dissociation constant, \( K_{a} \), defined as:

\[ K_{a} = \frac{[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COO}^{-}][\mathrm{H}^+]}{[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{COOH}]} \]This expression is key in calculating \( K_{a} \) with known concentrations.

Concentration Calculation

To find the acid dissociation constant \( K_{a} \) for the weak acid, lactic acid, we use the concentrations derived from the equilibrium expression. Given that the dissociation is small for lactic acid when the pH is 2.44, the value of \( x \) (which represents the change in concentration) is minimal compared to the initial concentration.

Therefore, we can simplify the expression by assuming \( x \ll 0.10 \), which allows us to express \( K_{a} \) as:

\[ K_{a} = \frac{(x)(x + 3.6 \times 10^{-3})}{0.10} \]

Using the determined \( [\mathrm{H}^+] \approx 3.6 \times 10^{-3} \) M, we plug the value into the equation to find:

• \( K_{a} = \frac{(3.6 \times 10^{-3})(3.6 \times 10^{-3})}{0.10} \)

This results in \( K_{a} \approx 1.3 \times 10^{-4} \).

Understanding the minor changes in concentration and simplifying assumptions are crucial in acquiring the correct \( K_{a} \) value, showcasing the importance of precision in chemical calculations.