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Lactic acid (CH \(_{3} \mathrm{CHOHCO}_{2} \mathrm{H}\) ) is found in sour milk, in sauerkraut, and in muscles after activity (see page 479 ). \(\left(K_{\mathrm{a}} \text { for lactic acid }=1.4 \times 10^{-4} .\right)\) (a) If 2.75 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 final pH lower or higher than the pH of the lactic acid solution?

Short Answer

Expert verified
The pH is 3.54, higher than the initial lactic acid solution.

Step by step solution

01

Calculate Moles of Sodium Lactate

Given that the molar mass of sodium lactate (NaCH₃CHOHCO₂) is approximately 112.06 g/mol, calculate the number of moles of sodium lactate using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).\[ \text{moles of sodium lactate} = \frac{2.75 \text{ g}}{112.06 \text{ g/mol}} \approx 0.0245 \text{ mol} \]
02

Calculate Concentration of Sodium Lactate in Solution

The total volume of the buffer solution is 500 mL (or 0.500 L). Calculate the concentration of sodium lactate in this solution:\[ [\text{NaCH}_3\text{CHOHCO}_2] = \frac{0.0245 \text{ mol}}{0.500 \text{ L}} = 0.049 \text{ M} \]
03

Use Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for buffer solutions is given by:\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Where \( \text{pK}_a = -\log(1.4 \times 10^{-4}) \approx 3.85 \). \([\text{A}^-]\) is the concentration of sodium lactate and \([\text{HA]}\) is the concentration of lactic acid (0.100 M).
04

Calculate pH of Buffer Solution

Substitute the known values into the Henderson-Hasselbalch equation:\[ \text{pH} = 3.85 + \log \left( \frac{0.049}{0.100} \right) \]\[ \text{pH} \approx 3.85 + \log(0.49) \approx 3.85 - 0.31 \approx 3.54 \]
05

Determine Change in pH

A \( 0.100 \text{ M} \) lactic acid solution without sodium lactate would have a pH of a typical weak acid around the pKa: approximately 2.0. Since 3.54 is higher than 2.0, the pH has increased.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a simple formula used in chemistry to estimate the pH of a buffer solution. Buffer solutions are important because they resist changes in pH when small amounts of acids or bases are added. This equation relates the pH with the concentrations of acid and its conjugate base in the solution. It is written as: , where [\text{A}^-] is the concentration of the conjugate base and [\text{HA}] is the concentration of the acid.
In the context of the exercise, this equation helps us understand how the pH of a lactic acid solution is affected by adding sodium lactate.
  • \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant \(K_a\), which measures the strength of the acid.
  • By substituting the concentrations of sodium lactate and lactic acid into the equation, we can calculate the pH of the buffer solution formed.
This particular example involves lactic acid (CH₃CHOHCO₂H) acting as the acid, and sodium lactate (its conjugate base) being added to form the buffer. Understanding this equation can help predict how changes in concentrations will affect the pH.
Lactic Acid
Lactic acid is an organic compound with the formula CH₃CHOHCO₂H, known for its role in muscle soreness after intense exercise and its presence in fermented foods like yogurt and sauerkraut. It's a weak acid with a dissociation constant \(K_a\) of \(1.4 \times 10^{-4}\), meaning it doesn’t completely ionize in water. This partial ionization is what classifies it as a weak acid.
When you dissolve lactic acid in water, some of the molecules dissociate to release hydrogen ions (H⁺) and lactate ions (C₃H₅O₃⁻). In a buffer, it’s these ions that help resist changes in pH.
  • Lactic acid’s low pKa value (around 3.85) suggests that it loses its proton more easily compared to stronger acids, but less easily than very weak acids.
  • Because lactic acid is common in traditional food items and biological processes, understanding its chemical behavior is useful in everyday contexts.
The presence of lactic acid in a buffer solution is crucial because it pairs with its conjugate base, lactate, to maintain a stable pH. This dynamic balance is essential in biological systems and various industrial applications.
Sodium Lactate
Sodium lactate is the sodium salt of lactic acid. Chemically represented as \(\text{NaCH}_3\text{CHOHCO}_2\), it is commonly used in food and pharmaceutical industries as a preservative and pH regulator.

Its role in forming a buffer solution is significant. In the buffer, sodium lactate acts as the conjugate base, which helps balance pH by reacting with excess hydrogen ions. When sodium lactate is added to lactic acid:
  • It increases the concentration of the conjugate base (lactate ions).
  • These ions are crucial to neutralizing added acids, thus maintaining the buffer’s pH.
  • The presence of sodium ions instead of hydrogen ions in its structure makes it less acidic than lactic acid itself.
By calculating the concentration of sodium lactate in the solution, you can then use it in the Henderson-Hasselbalch equation to find the pH, as shown in the exercise.
Understanding sodium lactate’s role aids in grasping how buffer solutions work and their importance in stabilizing environments against pH fluctuations.

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Most popular questions from this chapter

Which of the following barium salts should dissolve in a strong acid such as HCl: \(\mathrm{Ba}(\mathrm{OH})_{2}, \mathrm{BaSO}_{4},\) or \(\mathrm{BaCO}_{3} ?\)

Does the pH of the solution increase, decrease, or stay the same when you (a) Add solid ammonium chloride to a dilute aqueous solution of \(\mathrm{NH}_{3} ?\) (b) Add solid sodium acetate to a dilute aqueous solution of acetic acid? (c) Add solid NaCl to a dilute aqueous solution of NaOH?

Solid silver iodide, AgI, can be dissolved by adding aqueous sodium cyanide to it. $$\mathrm{AgI}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq}) \rightleftarrows\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq})$$ Show that this equation is the sum of two other equations, one for dissolving AgI to give its ions and the other for the formation of the \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) ion from \(\mathrm{Ag}^{+}\) and \(\mathrm{CN}^{-}\). Calculate \(K_{\text {net }}\) for the overall reaction.

What is the pH of the solution that results from adding \(25.0 \mathrm{mL}\) of \(0.12 \mathrm{M} \mathrm{HCl}\) to \(25.0 \mathrm{mL}\) of \(0.43 \mathrm{M} \mathrm{NH}_{3} ?\)

At \(20^{\circ} \mathrm{C},\) a saturated aqueous solution of silver acetate, \(\mathrm{AgCH}_{3} \mathrm{CO}_{2},\) contains \(1.0 \mathrm{g}\) of the silver compound dissolved in \(100.0 \mathrm{mL}\) of solution. Calculate \(K_{\mathrm{sp}}\) for silver acetate. $$\mathrm{AgCH}_{3} \mathrm{CO}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})$$

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