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Lactic acid \(\left(K_{\mathrm{a}}=1.4 \times 10^{-4}\right)\) and pyruvic acid \(\left(K_{\mathrm{a}}=3.2 \times 10^{-3}\right)\) are very important in human metabolism. Calculate the [conjugate acid]/[conjugate base] ratio for each such that the two acid samples would have the same \(\mathrm{pH}\). Explain your results.

Short Answer

Expert verified
Lactic acid: 0.71; Pyruvic acid: 0.031

Step by step solution

01

Understanding the Problem

We need to find the ratio \([\text{conjugate acid}]/[\text{conjugate base}]\) for both lactic and pyruvic acid, so that they have the same pH.
02

Using the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \( \text{pH} = \text{pK}_a + \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) \). Rearranging gives: \( \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) = \text{pH} - \text{pK}_a \).
03

Calculate pKₐ

For lactic acid: \( \text{pK}_{a,\text{lactic}} = -\log(1.4 \times 10^{-4}) \approx 3.85 \). For pyruvic acid: \( \text{pK}_{a,\text{pyruvic}} = -\log(3.2 \times 10^{-3}) \approx 2.49 \).
04

Setting Up Equal pH Conditions

Since both acids have the same pH, we write: \( \text{pH}_{\text{lactic}} = \text{pH}_{\text{pyruvic}} \).
05

Calculate the [Acid]/[Base] Ratio for Lactic Acid

Set \( \text{pH} = \text{pK}_{a,\text{lactic}} + \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) \). If we assume \( \text{pH} = 4 \), then \( 4 = 3.85 + \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) \). Solving gives: \( \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) = 0.15 \). \( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} = 10^{0.15} \approx 1.41 \). Thus, \( \frac{[\text{conjugate acid}]}{[\text{conjugate base}]} \approx 0.71 \).
06

Calculate the [Acid]/[Base] Ratio for Pyruvic Acid

Set \( \text{pH} = \text{pK}_{a,\text{pyruvic}} + \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) \). If we assume \( \text{pH} = 4 \), then \( 4 = 2.49 + \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) \). Solving gives: \( \log\left( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} \right) = 1.51 \). \( \frac{[\text{conjugate base}]}{[\text{conjugate acid}]} = 10^{1.51} \approx 32.4 \). Thus, \( \frac{[\text{conjugate acid}]}{[\text{conjugate base}]} \approx 0.031 \).
07

Summary

For lactic acid, the \([\text{conjugate acid}]/[\text{conjugate base}]\) ratio is approximately 0.71. For pyruvic acid, it is approximately 0.031 to achieve the same pH.

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Key Concepts

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

Acid-Base Chemistry
In the realm of acid-base chemistry, understanding how acids and bases interact in solutions is crucial. This field investigates the properties and behaviors of acids, which donate protons (H⁺), and bases, which accept protons.

Acid-base reactions are essential in many biological systems, industrial processes, and environmental contexts. Two significant concepts to grasp are:
  • Acids: Substances that can donate a proton to another substance. Examples include hydrochloric acid (HCl) and sulfuric acid (H₂SO₄).
  • Bases: Substances that can accept a proton. Common bases are sodium hydroxide (NaOH) and ammonia (NH₃).
The strength of an acid is often depicted by its dissociation constant, denoted as \(K_a\). The higher the \(K_a\) value, the stronger the acid, meaning it dissociates more in solution to release protons.

In this exercise, lactic acid and pyruvic acid are evaluated, both of which are important acids in metabolism. Their \(K_a\) values help determine their strength and how they will behave in solutions. It's essential to note that many biological systems use the balance of acids and bases to maintain homeostasis.
pH Calculation
The calculation of pH is fundamental in understanding how acidic or basic a solution is. pH is a measure of the hydrogen ion concentration in a solution, represented as:
\[ ext{pH} = - ext{log}_{10}[ ext{H}^+] \]

A lower pH means a more acidic solution, while a higher pH indicates a more basic or alkaline one. For neutral solutions, the pH is around 7. Here's how pH relates to acidity and basicity:
  • Acidic Solutions: Have a pH less than 7.
  • Neutral Solutions: Have a pH equal to 7.
  • Basic Solutions: Have a pH greater than 7.

In the exercise, we use the Henderson-Hasselbalch equation to relate pH with \(pK_a\) and the ratio of conjugate base to acid. This equation provides a vital connection between the concentration of compounds and pH, allowing us to calculate the conditions needed for equilibrium in a buffer system.
Conjugate Acid-Base Pairs
A conjugate acid-base pair consists of two species related by the gain or loss of a proton. When an acid donates a proton, the remaining entity is its conjugate base. Conversely, when a base gains a proton, it forms its conjugate acid.

Conjugate pairs are central to understanding how buffer solutions work. These solutions resist drastic changes in pH upon the addition of small amounts of acids or bases. In the Henderson-Hasselbalch equation, the presence of conjugate acid-base pairs is crucial. It is represented by:
\[ ext{pH} = ext{pK}_a + ext{log} rac{[ ext{conjugate base}]}{[ ext{conjugate acid}]} \]

Understanding these pairs helps explain the behavior of acids and bases when they are in solution. For lactic and pyruvic acids, their ability to form conjugate pairs forms the basis of calculating the \(pH\) and predicting how different concentrations of acid and base can affect the solution's acidity.
  • Conjugate Acid: Species formed when a base gains a proton.
  • Conjugate Base: Species remaining after an acid has donated a proton.
The exercise demonstrates using these pairs to maintain a stable pH environment, which is vital in many biochemical pathways and industrial processes.

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Most popular questions from this chapter

$$ \begin{aligned} &\text { Calculate the solubility }(\mathrm{mol} / \mathrm{L}) \text { of } \mathrm{SrSO}_{4}\\\ &\left(K_{\mathrm{sp}}=3.2 \times 10^{-7}\right) \text {in } 0.010-\mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4} \end{aligned} $$

A 30.00 -mL solution of 0.100 -M benzoic acid, a monoprotic acid, is titrated with \(0.100-\mathrm{M} \mathrm{NaOH}\). The \(K_{\mathrm{a}}\) of benzoic acid is \(1.2 \times 10^{-4}\). Determine the \(\mathrm{pH}\) after each of these volumes of titrant has been added: (a) \(10.00 \mathrm{~mL}\) (b) \(30.00 \mathrm{~mL}\) (c) \(40.00 \mathrm{~mL}\)

The grid has six lettered boxes, each of which contains an item that may be used to answer the questions that follow. Items may be used more than once and there may be more than one correct item in response to a question. $$ \begin{array}{|l|l|l|} \hline \mathbf{A} & \mathbf{B} & \mathbf{C} \\ {\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}} & Q>K_{\mathrm{sp}} & \mathrm{NaOH} \\ \hline \mathbf{D} & \mathbf{E} & \mathbf{F} \\ \text { Decreasing } \mathrm{pH} & \mathrm{pH}=\mathrm{p} K_{\mathrm{a}} & \begin{array}{l} \text { Sufficient } \mathrm{KCl} \\ \text { added to } \mathrm{CuCl}(\mathrm{aq}) \end{array} \\ \hline \end{array} $$ Place the letter(s) of the correct selection(s) on the appropriate line. (a) Could dissolve \(\mathrm{Zn}(\mathrm{OH})_{2}\) _________ (b) Could be used to prepare a buffer from \(\mathrm{HPO}_{4}^{2-}\) ___________ (c) Halfway to the equivalence point in the titration of a weak, monoprotic acid with strong base __________ (d) \(\mathrm{SO}_{2}\) -related atmospheric phenomenon _____________ (e) Species formed by a Lewis acid-base reaction ______________ (f) General condition required for precipitation to occur __________ (g) [conj. base] \(=[\) conj. acid \(]\) in a buffer ___________

Describe what a complex ion is and give an example.

The Mohr titration has been used to determine chloride concentration in a sample. A standardized silver nitrate solution is used to precipitate chloride as silver chloride from the test solution. Potassium chromate is used as the indicator because it reacts with slight excess silver ion to form silver chromate, \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\), a red precipitate. In practice, the silver chromate starts to precipitate just after the silver chloride precipitation is complete because at the equivalence point the only source of silver ions is from the \(\mathrm{AgCl}\) precipitate dissociation. A \(0.2651-\mathrm{g}\) solid sample containing chloride was dissolved in sufficient water to form \(20.00 \mathrm{~mL}\) of solution. Potassium chromate indicator was added to the solution and the titration required \(31.32 \mathrm{~mL}\) of standardized \(0.1041 \mathrm{M} \mathrm{AgNO}_{3} .\) (a) Calculate the mass percent chloride in the sample. (b) In a separate experiment it was determined that \(20.00 \mathrm{~mL}\) of the potassium chromate indicator solution required \(0.03 \mathrm{~mL}\) of the standardized \(\mathrm{AgNO}_{3}\) solution. Use this information to recalculate the mass percent chloride in the solid sample. \(K\) sp values: \(\mathrm{AgCl}=1.8 \times 10^{-10} ; \mathrm{Ag}_{2} \mathrm{CrO}_{4}=1.1 \times 10^{-12}\)

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