Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 160

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. $C_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}$ \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

Short Answer

Expert verified
The ranked solutions in order of increasing pH are: a. HI < HF < NaF < NaI b. HBr < NH\(_{4}\)Br < KBr < NH\(_{3}\) c. HNO\(_{3}\) < HOC\(_{6}\)H\(_{5}\) < C\(_{6}\)H\(_{5}\)NH\(_{3}\)NO\(_{3}\) < NaNO\(_{3}\) < KOC\(_{6}\)H\(_{5}\) < C\(_{6}\)H\(_{5}\)NH\(_{2}\) < NaOH

Step by step solution

01

Identify the acidity/basicity of each solution

HI and HF are strong acids, so they will have low pH values. NaF and NaI are salts formed from neutralizing the strong acids with a strong base, so their solutions will be slightly basic due to the hydrolysis of F- and I-, which will have pH values greater than 7.
02

Rank the solutions

We can rank the solutions as follows: HI < HF < NaF < NaI Solution b:
03

Identify the acidity/basicity of each solution

NH\(_{4}\)Br is a weak acidic solution, as it is formed from NH\(_{3}\), a weak base, and a strong acid, HBr. HBr is a strong acid. KBr is a salt formed from a strong acid and a strong base, so it will be neutral. NH\(_{3}\) is a weak base.
04

Rank the solutions

We can rank the solutions as follows: HBr < NH\(_{4}\)Br < KBr < NH\(_{3}\) Solution c:
05

Identify the acidity/basicity of each solution

C\(_{6}\)H\(_{5}\)NH\(_{3}\)NO\(_{3}\) is formed from a weak base and a strong acid, so it will be a weak acidic solution. NaNO\(_{3}\) is a salt formed from a strong acid and a strong base, so it will be neutral. NaOH is a strong base. HOC\(_{6}\)H\(_{5}\) is a weak acid. KOC\(_{6}\)H\(_{5}\) is a salt formed from a weak acid and a strong base, so it will be a weak basic solution. C\(_{6}\)H\(_{5}\)NH\(_{2}\) is a weak base. HNO\(_{3}\) is a strong acid.
06

Rank the solutions

We can rank the solutions as follows: HNO\(_{3}\) < HOC\(_{6}\)H\(_{5}\) < C\(_{6}\)H\(_{5}\)NH\(_{3}\)NO\(_{3}\) < NaNO\(_{3}\) < KOC\(_{6}\)H\(_{5}\) < C\(_{6}\)H\(_{5}\)NH\(_{2}\) < 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

Place the species in each of the following groups in order of increasing acid strength. a. $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\( (bond energies: \)\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}$ $\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )$ b. $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}$ c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: $\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},$ 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasant tart taste. Assuming that in cola drinks the concentration of phosphoric acid is \(0.007 M,\) calculate the \(\mathrm{pH}\) of this solution.

A typical sample of vinegar has a pH of \(3.0 .\) Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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