Problem 8 - Chemistry and Chemical Reactivity

Question

Lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CHOHCO}_{2} \mathrm{H}\right)\) is found in sour milk, in sauerkraut, and in muscles after activity. \(\left(K_{\mathrm{a}} \text { for lactic acid }=1.4 \times 10^{-4} .\right)\) (a) If \(2.75 \mathrm{g}\) of \(\mathrm{NaCH}_{3} \mathrm{CHOHCO}_{2}\), sodium lactate, is added to \(5.00 \times 10^{2} \mathrm{mL}\) of \(0.100 \mathrm{M}\) lactic acid, what is the \(\mathrm{pH}\) of the resulting buffer solution? (b) Is the \(\mathrm{pH}\) of the buffered solution lower or higher than the \(\mathrm{pH}\) of the lactic acid solution?

Step-by-Step Solution

Answer

(a) The pH is 3.53. (b) The pH is lower than the lactic acid solution.

Step 1: Calculate Moles of Sodium Lactate

We start by calculating the number of moles of sodium lactate. Given the mass is 2.75 g, and the molar mass of sodium lactate (\( \text{NaCH}_3\text{CHOHCO}_2 \)) is approximately 112.06 g/mol, we calculate the moles using the formula: \( \text{Moles} = \text{Mass} / \text{Molar Mass} \). Thus, the moles are: \( \frac{2.75\, \text{g}}{112.06\, \text{g/mol}} = 0.0245 \text{ moles} \).

Step 2: Determine the Concentration of Sodium Lactate

Next, we find the concentration of sodium lactate in the 500 mL solution. The concentration \([\text{C}]\) is given by \([\text{C}] = \frac{\text{Moles}}{\text{Volume in L}}\). Using the volume of 0.500 L, the concentration is: \( \frac{0.0245 \text{ moles}}{0.500 \text{ L}} = 0.049 \text{ M} \).

Step 3: Calculate the Initial Concentration of Lactic Acid

The initial concentration of lactic acid is given directly as 0.100 M for a volume of 500 mL (0.500 L).

Step 4: Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where \([\text{A}^-]\) is the concentration of the sodium lactate (0.049 M) and \([\text{HA}]\) is the concentration of lactic acid (0.100 M). First, calculate \( \text{pK}_a \) as \( \text{pK}_a = -\log(1.4 \times 10^{-4}) \approx 3.85 \). Substitute these values into the equation: \( \text{pH} = 3.85 + \log \left( \frac{0.049}{0.100} \right) \approx 3.85 + (-0.318) = 3.53 \).

Step 5: Compare pH Values

The pH of the lactic acid solution alone, without buffer, would be lower than 3.85, given by its Ka value. The observed pH of the buffer solution is 3.53, indicating it is lower than the pH of the lactic acid solution without the salt.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a powerful tool in chemistry for calculating the pH of a buffer solution. A buffer solution contains a weak acid and its conjugate base. This equation is expressed as:

\[\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]Here, \([\text{A}^-]\) is the concentration of the base (like sodium lactate), and \([\text{HA}]\) is the concentration of the acid (such as lactic acid).

The term \(\text{pKa}\) is the negative logarithm of the acid dissociation constant \((K_a)\). It provides insight into the acid strength; a lower \(\text{pKa}\) means a stronger acid. By using the concentrations of both acid and its conjugate base, this formula helps determine the pH balance maintained by the buffer.

Acid-Base Equilibria

Acid-base equilibria are foundational in understanding how solutions maintain specific pH levels. In a typical acid-base reaction, an acid donates a proton to a base. The equilibrium is established between the ionized (or dissociated) and non-ionized forms of the acid and base.

In a buffer, you'll find an acid \((HA)\) and its conjugate base \((A^-)\). The equilibrium can be represented as:
\[\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\]Buffers resist changes in pH when small amounts of acid or base are added. This property makes them essential in biological and chemical processes where pH stability is crucial.

Sodium Lactate

Sodium lactate is the salt form of lactic acid, playing a crucial role in buffer solutions. Being a conjugate base, sodium lactate comes into play when lactic acid loses a proton. In the formation of a buffer, sodium lactate reacts with lactic acid to stabilize the pH of the solution.

In our exercise, we calculated the moles of sodium lactate using its mass and molar mass. This helps in determining its concentration in solution, which is a vital step for using the Henderson-Hasselbalch equation effectively. Sodium lactate, through its buffering action, helps to resist changes in pH by reacting with any added acids or bases.

Lactic Acid

Lactic acid is a common organic acid, known scientifically as \(\text{CH}_3\text{CHOHCO}_2\text{H}\). It is a weak acid often found in sour milk and muscles after activity. Its strength is indicated by the acid dissociation constant, \(K_a\), which measures how well the acid donates protons.

For lactic acid, \(K_a = 1.4 \times 10^{-4}\), a low value showing it is a weak acid. The acid's ability to donate protons partially makes it perfect for forming buffer solutions with its conjugate base, sodium lactate. In this buffer system, lactic acid donates protons to neutralize added bases, while sodium lactate can neutralize added acids, thus maintaining a relatively stable pH.