Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the pH of the solution related to the \(\mathrm{p} K_{\mathrm{a}}\) value of the weak acid? If [conjugate base \(]>[\text { weak acid }]\) , how is pH related to \(\mathrm{p} K_{\mathrm{a}} ?\)

Short Answer

Expert verified
In a buffer solution, the pH is related to the pKa value of the weak acid following the Henderson-Hasselbalch equation: \( pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \). If [weak acid] > [conjugate base], the fraction inside the log is less than 1, resulting in a negative log value and a pH lower than the pKa. Conversely, if [conjugate base] > [weak acid], the fraction inside the log is greater than 1, leading to a positive log value and a pH higher than the pKa.

Step by step solution

01

Understand the Buffer Solution

A buffer solution is an aqueous solution that consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. The purpose of a buffer solution is to maintain a relatively constant pH when a small amount of acid or base is added to it.
02

Recall the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a way to determine the pH of a buffer solution. The equation is given by: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \]
03

Find the relationship between pH and pKa for [weak acid] > [conjugate base]

Using the Henderson-Hasselbalch equation: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \] Since [weak acid] > [conjugate base], the fraction inside the log will be less than 1. The log of a number smaller than 1 is negative, so the pH will be lower than the pKa in this scenario.
04

Find the relationship between pH and pKa for [conjugate base] > [weak acid]

Again, using the Henderson-Hasselbalch equation: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \] In this case, since [conjugate base] > [weak acid], the fraction inside the log will be greater than 1. The log of a number greater than 1 is positive, so the pH will be higher than the pKa in this situation. In conclusion, if [weak acid] > [conjugate base], the pH of the buffer solution will be lower than the pKa value of the weak acid, while if [conjugate base] > [weak acid], the pH of the buffer solution will be higher than the pKa value of the weak acid.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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 fundamental tool in understanding the pH of buffer solutions. It provides a mathematical relationship between the pH, the pKa, and the concentrations of the acid and its conjugate base. The equation is expressed as: \[ pH = pKa + \log \left(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\right) \]This formula allows us to predict how the pH will adjust if either the acid or base concentration changes. Here,
  • pH represents the acidity level of the solution.
  • pKa is a constant that indicates the strength of the weak acid. It represents the pH at which half of the acid is dissociated.
  • The fraction \(\frac{[\text{conjugate base}]}{[\text{weak acid}]}\) shows the ratio of the concentrations of the base form to the acid form.
Using this equation, one can determine how strong the buffer solution is at maintaining a steady pH, even when small amounts of acids or bases are added.
Weak acid and conjugate base
In chemistry, buffers are solutions that resist changes in pH. They are composed of a weak acid and its conjugate base. What are weak acids?
Weak acids partially dissociate in solution. This means they do not completely ionize into ions in water. Their dissociation is reversible and the equilibrium lies to the left, meaning most of the acid remains undissociated. Examples include acetic acid and citric acid.Understanding conjugate bases
A conjugate base is what remains of the acid after it donates a proton (H\(^{+}\)). For example, when acetic acid (CH\(_3\)COOH) loses a proton, it forms its conjugate base, acetate (CH\(_3\)COO\(^{-}\)).
  • The weak acid equilibrates with its conjugate base, forming a buffer.
  • This buffer solution can absorb excessive H\(^{+}\) or OH\(^{-}\) ions, preventing significant changes in pH.
The balance between the weak acid and its conjugate base is crucial. If more conjugate base exists, it can neutralize added acids. Conversely, more weak acid can neutralize added bases.
pKa and pH relationship
The concepts of pKa and pH are intimately related and pivotal in understanding buffer systems. The pKa is the negative logarithm of the acid dissociation constant (Ka) and gives insight into the strength of an acid. Understanding the relationship
  • The pKa value tells us the pH at which an acid is half dissociated, meaning half of the acid molecules have donated a proton.
  • In a buffer, comparing the pH with the pKa allows us to predict the behavior of the solution. A pH lower than the pKa indicates a higher concentration of the weak acid than its conjugate base, while a pH higher than the pKa suggests more conjugate base than weak acid.
This relationship is critical for designing buffers for specific pH levels. For example, to prepare a buffer with a pH close to 7.4 (the physiological pH), one would select an acid with a pKa near this value. This ensures the buffer can effectively resist changes and maintain the desired pH even with minor additions of acids or bases.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the volume of \(1.50 \times 10^{-2} M \mathrm{NaOH}\) that must be added to 500.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15 .\)

The active ingredient in aspirin is acetylsalicylic acid. A 2.51 -g sample of acetylsalicylic acid required 27.36 \(\mathrm{mL}\) of 0.5106 \(\mathrm{M} \mathrm{daOH}\) for complete reaction. Addition of 13.68 \(\mathrm{mL}\) of 0.5106\(M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH \(=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{2}\) value. State any assumptions you must make to reach your answer.

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HONH}_{2}\) by 0.100 \(\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)\) a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HCO}_{3}\) - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free