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

Use of Molar Concentrations to Calculate \(\mathrm{pHI}\) What is the \(\mathrm{pH}\) of a solution that contains \(0.20 \mathrm{~m}\) sodium acetate and \(0.60\) m acetic acid \(\left(\mathrm{p} K_{\mathrm{n}}=4.76\right)\) ?

Short Answer

Expert verified
The pH of the solution is approximately 4.28.

Step by step solution

01

Identify Components

Determine which components of the solution will affect the pH. The solution contains acetic acid (a weak acid) and sodium acetate (its conjugate base). This is a buffer solution.
02

Apply Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution: \[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]Where \([A^-]\) is the concentration of acetate ions and \([HA]\) is the concentration of acetic acid.
03

Substitute Values

Substitute the given concentrations and the \( pK_a \) value into the Henderson-Hasselbalch equation:\[pH = 4.76 + \log\left(\frac{0.20}{0.60}\right)\]
04

Calculate Ratio

Calculate the ratio of sodium acetate to acetic acid: \[ \log\left(\frac{0.20}{0.60}\right) = \log\left(\frac{1}{3}\right) \approx -0.477 \]
05

Compute the pH

Add the calculated value of the logarithmic ratio to the \( pK_a \):\[ 4.76 - 0.477 = 4.28 \]Thus, the pH of the solution is approximately 4.28.

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 an important formula used to calculate the pH of a buffer solution, which consists of a weak acid and its conjugate base. Although it may seem complex at first, it's actually quite straightforward: \[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]Here:
  • \([A^-]\) represents the concentration of the conjugate base, which is the anionic part of the weak acid after it donates a proton.
  • \([HA]\) symbolizes the concentration of the weak acid itself.
  • \(pK_a\) is the acid dissociation constant, indicating the strength of the weak acid in aqueous solution. In this case, it's 4.76 for acetic acid.
The equation is derived from the equilibrium constant expression for a weak acid. It provides a simple method to estimate the pH based on the ratio of these concentrations. Understanding this equation is essential for managing buffer systems in various chemical and biological applications.
buffer solution
A buffer solution is a special type of solution designed to resist significant changes in pH upon the addition of small amounts of acids or bases. The key to maintaining this stability is its composition:
  • It includes a weak acid and its conjugate base in appreciable amounts.
  • This combination allows the solution to neutralize added hydroxide ions or protons, thereby maintaining a nearly constant pH.
In our example, the buffer consists of acetic acid (weak acid) and sodium acetate (conjugate base). This system can counteract pH changes effectively, making buffer solutions crucial in many chemical reactions and biological systems, like blood, where maintaining a specific pH range is vital. It's this balancing act of neutralizing acids and bases that defines the functionality and utility of buffer solutions.
acetic acid
Acetic acid is a common weak acid, often recognized as the main component of vinegar. Its chemical formula is \(CH_3COOH\), and it's known by its characteristic sour taste and pungent smell.
  • Being a weak acid, it does not completely dissociate in water, meaning not all acetic acid molecules donate protons to form acetate ions \(CH_3COO^-\) and hydronium ions \(H_3O^+\).
  • The constant \(pK_a\) for acetic acid is 4.76, which indicates its acid strength.
In buffer solutions, the acetic acid part can donate protons to neutralize added bases, which aligns perfectly with our exercise of calculating the pH of buffer solutions. Understanding acetic acid's behavior in solutions is important for grasping how buffer systems work in neutralizing added bases.
sodium acetate
Sodium acetate is the salt of acetic acid and is often used as the conjugate base component in buffer solutions. It forms when acetic acid is neutralized by sodium hydroxide. Here's what you should know about it:
  • Its chemical formula is \(CH_3COONa\), indicating that it's composed of sodium ions \(Na^+\) and acetate ions \(CH_3COO^-\).
  • In solution, sodium acetate dissociates completely, providing a steady supply of acetate ions.
  • These acetate ions play a crucial role in buffering, as they can react with any added protons, mitigating pH changes.
In the pH calculation scenario, the role of sodium acetate is vital as it contributes the conjugate base \([A^-]\) in the Henderson-Hasselbalch equation. This ensures that the system can respond to added acids, confirming the buffer's effectiveness in maintaining a stable pH level.

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

Properties of Strong and Weak Acids Classify each acid or property as representing a strong acid or a weak acid: a. hydrochloric acid; b. acetic acid; c. strong tendency to dissociate protons; d. larger \(K_{3}\); e. partially dissociates into ions; f. larger \(\mathrm{p} K_{\mathrm{a}^{*}}\)

Calculation of the \(\mathrm{pH}\) of a Strong Acid or Base a. Write out the acid dissociation reaction for hydrochloric acid. b. Calculate the \(\mathrm{pH}\) of a solution of \(5 \times 10^{-4} \mathrm{M}\) hydrochloric acid at \(25^{\circ} \mathrm{C}\). c. Write out the acid dissociation reaction for sodium hydroxide. d. Calculate the \(\mathrm{pH}\) of a solution of \(7 \times 10^{-5} \mathrm{M}\) sodium hydroxide at \(25^{\circ} \mathrm{C}\).

\- Control of Blood pll by Respiratory Rate a. The partial pressure of \(\mathrm{CO}_{2}\left(\mathrm{~T} \mathrm{CO}_{2}\right)\) in the lungs can be varied rapadly by the rate and depth of breathing. For example, a common remedy to alleviate hiccups is to increase the concentration of \(\mathrm{CO}_{2}\) in the lungs. This can be achieved by holding one's breath, by very slow and shallow breathing (hypoventilation), or by breathing in and out of a paper bag. Under such conditions, \(\mathrm{p} \mathrm{CO}_{2}\) in the air space of the lungs rises above normal. How would increasing \(\mathrm{pCO}_{2}\) in the air space of the lungs affect blood pH?b. It is common practice among competitive shortdistance runners to breathe rapidly and deeply (hyperventilate) for about half a minute to remove \(\mathrm{CO}_{2}\) from their lungs just before a race begins. Under these conditions, blood pH may rise to \(7.6\). Explain how hyperventilation elicits an increase in blood pH. c. During a short-distance run, the muscles produce a large amount of lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} ; K_{\mathrm{a}}=1.38 \times 10^{-4} \mathrm{M}\right)\) from their glucose stores. Why might hyperventilation before a short-distance run be useful?

Reological Effects of pH The defendant in a lawstait over industrial pollution is accused of releasing effluent of pHI 10 into a trout stream. The plaintiff has asked that the defendant reduce the eftluent's pI to no higher than 7 . The defendant's attorney, aiming to please the court, promises that his client will do even better than that: the defendant will bring the pH of the effluent down to 1! a. Will the defense attorney's suggested remecty be acceptable to the plaintiff? Why or why not? b. What facts about pH does the defense attorney need to understand?

Working with Buffers A buffer contains \(0.010\) mol of lactic acid \(\left(\mathrm{p} K_{\mathrm{n}}=3.56\right)\) and \(0.050\) mol of sodium lactate per liter. a. Calculate the pH of the buffer. b. Calculate the change in \(\mathrm{pIl}\) after adding \(5.0 \mathrm{~mL}\) of \(0.5\) M MCl to 1 L of the huffer. c. What pH change would you expect if you added the same quantity of HCl to 1 L of pure water?

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