Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 50

Calculate the \(\left[\mathrm{H}^{+}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C}\) . Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{OH}^{-}\right]=1.5 M\) b. \(\left[\mathrm{OH}^{-}\right]=3.6 \times 10^{-15} M\) c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\) d. \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-4} M\) Also calculate the pH and pOH of each of these solutions.

Short Answer

Expert verified
The calculated concentrations of \(\left[\mathrm{H}^{+}\right]\) and pH and pOH values are as follows: a. \(\left[\mathrm{H}^{+}\right] = 6.67 \times 10^{-15} M\), pH=14.18, pOH=-0.18, basic solution. b. \(\left[\mathrm{H}^{+}\right] = 2.77 \times 10^{-14} M\), pH=13.56, pOH=14.44, basic solution. c. \(\left[\mathrm{H}^{+}\right] = 1.0 \times 10^{-7} M\), pH=7, pOH=7, neutral solution. d. \(\left[\mathrm{H}^{+}\right] = 1.37 \times 10^{-12} M\), pH=11.86, pOH=3.14, basic solution.

Step by step solution

01

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Using the \(K_w\) equation, we can calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{1.5} = 6.67 \times 10^{-15} M\)
02

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(6.67 \times 10^{-15})}= 14.18\) \(pOH = -\log{(1.5)}= -0.18\) Since the pH value is greater than 7, it is a basic solution. b. \(\left[\mathrm{OH}^{-}\right]=3.6 \times 10^{-15} M\)
03

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Similar to part a, we will calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{3.60 \times 10^{-15}} = 2.77 \times 10^{-14} M\)
04

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(2.77 \times 10^{-14})}= 13.56\) \(pOH = -\log{(3.6 \times 10^{-15})}= 14.44\) Since the pH value is greater than 7, it is a basic solution. c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-7} M\)
05

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Similar to part a, we will calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{1 \times 10^{-7}} = 1.0 \times 10^{-7} M\)
06

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(1.0 \times 10^{-7})}= 7\) \(pOH = -\log{(1.0 \times 10^{-7})}= 7\) Since the pH value is equal to 7, it is a neutral solution. d. \(\left[\mathrm{OH}^{-}\right]=7.3 \times 10^{-4} M\)
07

Calculate the \(\left[\mathrm{H}^{+}\right]\)

Similar to part a, we will calculate the \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = \dfrac{K_w}{\left[\mathrm{OH}^{-}\right]} = \dfrac{1 \times 10^{-14}}{7.3 \times 10^{-4}} = 1.37 \times 10^{-12} M\)
08

Determine the pH and pOH

Now, we will calculate the pH and pOH values: \(pH = -\log{(1.37 \times 10^{-12})}= 11.86\) \(pOH = -\log{(7.3 \times 10^{-4})}= 3.14\) Since the pH value is greater than 7, it is a basic solution.

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

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C} ) .\) Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at $35.0^{\circ} \mathrm{C}$ is \(2.1 \times 10^{-14} .\)

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},$ assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

Arrange the following 0.10\(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$ \mathrm{CaBr}_{2}, \mathrm{KNO}_{2}, \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4} $$

Use Table 14.3 to help order the following bases from strongest to weakest. $$ \mathrm{NO}_{3}^{-}, \quad \mathrm{H}_{2} \mathrm{O}, \quad \mathrm{NH}_{3}, \quad \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} $$ Also order the following acids from strongest to weakest. $$ \mathrm{HNO}_{3}, \quad \mathrm{H}_{2} \mathrm{O}, \quad \mathrm{NH}_{4}^{+}, \quad \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} $$
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

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