Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 44

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 \(\mathrm{L}\) of each of the following solutions: a. 0.050\(M\) propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)\( and 0.080\)M$ sodium propanoate b. 0.50\(M\) propanoic acid and 0.80\(M\) sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Short Answer

Expert verified
a) The pH of solution a after adding NaOH is \(4.34\). b) The pH of solution b after adding NaOH is \(4.87\). c) Solution a does not act as a buffer after adding NaOH.

Step by step solution

01

Calculate the initial moles and concentration of each species

Before adding NaOH, we can calculate the initial moles and concentration of propanoic acid (HC3H5O2) and sodium propanoate (C3H5O2- Na+) in each solution. For solution a: Initial [HC3H5O2] = 0.050 M Initial [C3H5O2-] = 0.080 M For solution b: Initial [HC3H5O2] = 0.50 M Initial [C3H5O2-] = 0.80 M
02

Calculate the moles and concentration of species after NaOH is added

When NaOH is added, it reacts with propanoic acid as follows: HC3H5O2 + OH- -> C3H5O2- + H2O Since 0.15 mole of NaOH is added to 1.00 L of solution, the concentration of NaOH is 0.15 M. Now we will calculate the new concentration of propanoic acid and its conjugate base. For solution a: New [HC3H5O2] = 0.050 M - 0.15 M New [C3H5O2-] = 0.080 M + 0.15 M For solution b: New [HC3H5O2] = 0.50 M - 0.15 M New [C3H5O2-] = 0.80 M + 0.15 M
03

Calculate the pH using the Henderson-Hasselbalch equation

Now we can use the Henderson-Hasselbalch equation to find the pH of each solution: pH = pKa + log([C3H5O2-]/[HC3H5O2]) First, we need to determine pKa from the given Ka value: pKa = -log(Ka) = -log(1.3 × 10^{-5}) ≈ 4.89 Now we can find the pH for each solution: For solution a: pH = 4.89 + log((0.080 + 0.15)/(0.050 - 0.15)) ≈ 4.34 For solution b: pH = 4.89 + log((0.80 + 0.15)/(0.50 - 0.15)) ≈ 4.87
04

Determine if the solution in part a is still a buffer after adding NaOH

To check if solution a still acts as a buffer solution, we can look at the ratio of [C3H5O2-] to [HC3H5O2] after adding NaOH. The new concentrations are: [C3H5O2-] = 0.080 M + 0.15 M = 0.23 M [HC3H5O2] = 0.050 M - 0.15 M = -0.10 M Since the concentration of HC3H5O2 is negative after adding NaOH, it means that all of the weak acid has reacted with NaOH, and there is no weak acid left to maintain a buffer system. Therefore, solution a does not act as a buffer after adding NaOH. Final answer: a) The pH of solution a after adding NaOH is 4.34. b) The pH of solution b after adding NaOH is 4.87. c) Solution a does not act as a buffer after adding NaOH.

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

A buffer solution is prepared by mixing 75.0 \(\mathrm{mL}\) of 0.275 \(\mathrm{M}\) fluorobenzoic acid $\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{F}\right)\( with 55.0 \)\mathrm{mL}$ of 0.472 \(\mathrm{M}\) sodium fluorobenzoate. The \(\mathrm{pK}_{\mathrm{a}}\) of this weak acid is \(2.90 .\) What is the pH of the buffer solution?

Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M\) pyridine with 0.100\(M\) hydrochloric acid \(\left(K_{\mathrm{b}} \text { for pyridine is } 1.7 \times 10^{-9}\right) .\) Do not calculate the points at 24.9 and 25.1 \(\mathrm{mL}\)

Which of the following can be classified as buffer solutions? $$ \begin{array}{l}{\text { a. } 0.25 M \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}} \\ {\text { b. } 0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}} \\ {\text { c. } 0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{MKOCl}}\end{array} $$ $$ \begin{array}{l}{\text { d. } 0.70 M \mathrm{KOH}+0.70 \mathrm{M} \text { HONH_ }} \\ {\text { e. } 0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}}\end{array} $$

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{HCN}$ by 0.100 \(\mathrm{M} \mathrm{KOH}\) at $25^{\circ} \mathrm{C} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN}=6.2 \times 10^{-10} .\right)$ a. Calculate the pH after 0.0 \(\mathrm{mL}\) of KOH has been added. b. Calculate the pH after 50.0 \(\mathrm{mL}\) of KOH has been added. c. Calculate the pH after 75.0 \(\mathrm{mL}\) of KOH has been added. d. Calculate the pH at the equivalence point. e. Calculate the pH after 125 \(\mathrm{mL}\) of KOH has been added.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

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