Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 52

The \(\mathrm{pOH}\) of a sample of baking soda dissolved in water is \(5.74\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH},\left[\mathrm{H}^{+}\right]\), and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is the solution acidic or basic?

Short Answer

Expert verified
In summary, the solution has a pH of \(8.26\), a \(\mathrm{[H^{+}]}\) concentration equal to \(5.62 \times 10^{-9} \mathrm{M}\), and a \(\mathrm{[OH^{-}]}\) concentration equal to \(1.82 \times 10^{-6} \mathrm{M}\). The solution is basic.

Step by step solution

01

Find the pH from the given pOH

We know that at 25°C, the relationship between pH and pOH is: pH + pOH = 14 We're given that pOH = 5.74, so we can find the pH as follows: pH = 14 - 5.74 = 8.26
02

Calculate the concentration of H+ ions [H+]

Now that we have the pH, we can find the concentration of H+ ions using the definition of pH: pH = -log[H+] Rearrange the equation to find the H+ concentration: [H+] = 10^(-pH) = 10^(-8.26) = 5.62 × 10^(-9) M
03

Calculate the concentration of OH- ions [OH-]

We can find the concentration of OH- ions using the definition of pOH: pOH = -log[OH-] Rearrange the equation to find the OH- concentration: [OH-] = 10^(-pOH) = 10^(-5.74) = 1.82 × 10^(-6) M
04

Determine if the solution is acidic or basic

We can classify the solution as acidic, neutral, or basic based on its pH value: - If pH < 7, the solution is acidic - If pH = 7, the solution is neutral - If pH > 7, the solution is basic In our case, pH = 8.26, which is greater than 7. Therefore, the solution is basic. In summary, the solution has a pH of 8.26, a [H+] concentration equal to 5.62 × 10^(-9) M, and a [OH-] concentration equal to 1.82 × 10^(-6) M. The solution is basic.

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

Determine \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right]\), and the \(\mathrm{pH}\) of each of the following solutions. a. \(1.0 \mathrm{M} \mathrm{KCl}\) b. \(1.0 \mathrm{M} \mathrm{KF}\)

At \(25^{\circ} \mathrm{C}\), a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) has a pH of \(2.80\). Calculate the water solubility of benzoic acid in moles per liter.

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\mathrm{a}}\) value for hypochlorous acid is \(3.5 \times 10^{-8}\), which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) ?

A sample containing \(0.0500 \mathrm{~mol} \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make \(1.00 \mathrm{~L}\) of solution. This solution contains hydrated \(\mathrm{SO}_{4}{ }^{2-}\) and \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is \(6.73\) atm at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\). (To do this calculation, you must assume that none of the ions goes through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{~N}_{2} \mathrm{O}_{10} \mathrm{~S}\) as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise 143.) Why is codeine sulfate used instead of codeine?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

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