Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 88

You are asked to prepare a 100.0 mL sample of a solution with a pH of 5.50 by dissolving the appropriate amount of a solute in water with \(\mathrm{pH}=7.00 .\) Which of these solutes would you use, and in what quantity? Explain your choice. (a) \(15 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}) ;\) (b) \(12 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) (c) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) ;\) (d) glacial (pure) acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\)

Short Answer

Expert verified
To prepare a 100.0 mL solution with a pH of 5.50, use about 0.003 g of \(\mathrm{NH}_{4}\mathrm{Cl}\).

Step by step solution

01

Identify the nature of the solutes

Identify the nature of the solutes. Among them, only ammonium chloride would result in a solution with a pH less than 7, because it behaves as a weak acid when dissolved in water.
02

Determine the amount of \( \mathrm{NH}_{4} \mathrm{Cl} )

Use the formula: pH = pKa + log([A-]/[HA]), where A- is NH3 and HA is NH4+. Given that pKa of NH4+ is 9.25 and pH is 5.50, we can solve for the ratio of [A-]/[HA] using the above formula. This gives [A-]/[HA] = 10^(pH-pKa) = 10^(5.5 - 9.25) = 0.000562. The ratio is less than 1, which means that we need more HA, or NH4Cl, than A-, or NH3, to make the solution.
03

Calculate the concentration of NH3 and NH4Cl

Now, since NH3 is the solvent, we can assume its initial concentration to be 1M and NH4Cl to be xM. We can use the ratio obtained in step 2 to write: x / 1 = 0.000562. Solving for x, we get x = 1 * 0.000562 = 0.000562 M.
04

Calculate the amount of NH4Cl in grams

Now we can calculate the amount of NH4Cl in grams needed. First, convert the volume of solution required from milliliters to liters. 100 mL = 0.1 L. Then calculating the mass = Molarity * volume * molecular weight. For NH4Cl, Molarity = 0.000562 M, volume= 0.1 L, and molecular weight of NH4Cl = 53.5 g/mol. Solving the above equation gives the mass of NH4Cl in g = 0.000562 * 0.1 * 53.5 =~ 0.003 g.

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

Fluoroacetic acid occurs in gifblaar, one of the most poisonous of all plants. A 0.318 M solution of the acid is found to have a \(\mathrm{pH}=1.56 .\) Calculate \(K_{\mathrm{a}}\) of fluoroacetic acid. $$\mathrm{CH}_{2} \mathrm{FCOOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \quad\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CH}_{2} \mathrm{FCOO}^{-}(\mathrm{aq}) \quad K_{\mathrm{a}}=?$$

Codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{N},\) is an opiate, has analgesic and antidiarrheal properties, and is widely used. In water, codeine is a weak base. A handbook gives \(\mathrm{p} K_{\mathrm{a}}=6.05\) for protonated codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+} .\) Write the reaction for \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{O}_{3} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} K_{\mathrm{b}}\) for codeine.

The following four equilibria lie to the right: \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}+\) \(\mathrm{CH}_{3} \mathrm{NH}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ; \mathrm{H}_{2} \mathrm{SO}_{3}+\mathrm{F}^{-} \longrightarrow\) \(\mathrm{HSO}_{3}^{-}+\mathrm{HF} ; \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\) \(\mathrm{H}_{2} \mathrm{O} ;\) and \(\mathrm{HF}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{F}^{-}+\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) (a) Rank all the acids involved in order of decreasing acid strength. (b) Rank all the bases involved in order of decreasing base strength. (c) State whether each of the following two equilibria lies primarily to the right or to the left: (i) \(\mathrm{HF}+\mathrm{OH}^{-} \longrightarrow \mathrm{F}^{-}+\mathrm{H}_{2} \mathrm{O} ;\) (ii) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\) \(\mathrm{HSO}_{3}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{SO}_{3}\).

Indicate whether each of the following is a Lewis acid or base. (a) \(\mathrm{OH}^{-} ;\) (b) \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{B} ;\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH},\) and \(\mathrm{pOH}\) of \(0.55 \mathrm{M}\) \(\mathrm{M} \mathrm{HClO}_{2} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear 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