Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 14

Calculate the \(\mathrm{pH}\) of \(1.00 \mathrm{~L}\) of the buffer \(1.00 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COONa} / 1.00 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) before and after the addition of (a) \(0.080 \mathrm{~mol} \mathrm{NaOH}\) and (b) \(0.12 \mathrm{~mol}\) HCl. (Assume that there is no change in volume.)

Short Answer

Expert verified
The initial pH of the buffer solution is 4.74. After the addition of 0.080 mol of NaOH, the pH of the buffer solution is calculated as per Step 2. After the addition of 0.12 mol of HCl, the pH of the buffer solution is calculated as per Step 3. The actual numerical values will depend on the exact logarithmic values calculated.

Step by step solution

01

Calculate the initial pH

First, to calculate the initial pH of the buffer solution, one can use the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{pKa} + \log_{10} \left( \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right) \] Given that \(\mathrm{CH}_{3} \mathrm{COONa}\) is a base and \(\mathrm{CH}_{3} \mathrm{COOH}\) is an acid and both their concentrations are 1.00 M, the ratio of base to acid is 1. The pKa of acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) is 4.74. Therefore, the initial pH is \(4.74 + \log_{10}(1) = 4.74\).
02

Calculate the pH after the addition of NaOH

When NaOH is added to the solution, it reacts with the acetic acid, decreasing the concentration of acetic acid and increasing the concentration of acetate. Since 0.080 mol of NaOH is added, it reduces the concentration of acetic acid by 0.080 and increases the concentration of acetate by the same amount. Therefore the new concentrations are 0.92 M for acetic acid and 1.08 M for acetate. Substituting these values into the Henderson-Hasselbalch equation, \[ \mathrm{pH} = \mathrm{pKa} + \log_{10} \left( \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right) \] \[ \mathrm{pH} = 4.74 + \log_{10} \left( \frac{1.08}{0.92} \right) \]
03

Calculate the pH after the addition of HCl

When HCl is added to the solution, it reacts with the acetate, decreasing the concentration of acetate and increasing the concentration of acetic acid. Since 0.12 mol of HCl is added, it reduces the concentration of acetate by 0.12 and increases the concentration of acetic acid by the same amount. Therefore the new concentrations are 1.12 M for acetic acid and 0.88 M for acetate. Substituting these values into the Henderson-Hasselbalch equation, \[ \mathrm{pH} = \mathrm{pKa} + \log_{10} \left( \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right) \] \[ \mathrm{pH} = 4.74 + \log_{10} \left( \frac{0.88}{1.12} \right) \]

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.

Start your free trial

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

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

Specify which of these systems can be classified as a buffer system: (a) \(\mathrm{KCl} / \mathrm{HCl}\), (b) \(\mathrm{NH}_{3} / \mathrm{NH}_{4} \mathrm{NO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4} / \mathrm{NaH}_{2} \mathrm{PO}_{4}\)

The molar solubility of \(\mathrm{MnCO}_{3}\) is \(4.2 \times 10^{-6} \mathrm{M}\). What is \(K_{\mathrm{sp}}\) for this compound?

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a buret to an Erlenmeyer flask containing HCl solution and using phenolphthalein as indicator. At the equivalence point, he observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7 . Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).

Write the solubility product expression for the ionic compound \(\mathrm{A}_{x} \mathrm{~B}_{y}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation
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