Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 53

Is a solution that is \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}(\mathrm{aq})\) likely to be acidic, basic, or pH neutral? Explain.

Short Answer

Expert verified
The solution of 0.10M Na2S is expected to be basic since the sulfide ions react with water to produce hydroxide ions, causing an increase in the pH above 7.

Step by step solution

01

Dissolving Na2S in water

When Na2S dissolves in water, it dissociates into sodium ions (Na+) and sulfide ions (S2-), according to the reaction: \( \mathrm{Na}_{2} \mathrm{S}(aq) \rightarrow 2\mathrm{Na}^{+}(aq) + \mathrm{S}^{2-}(aq) \). The sodium ions do not interact with the water because they are the conjugates of a strong base (NaOH), and thus do not affect the acidity or basicity of the solution.
02

Reaction of sulfide ions with water

Sulfide ions (S2-) react with water to produce hydrogen sulfide (H2S) and hydroxide ions (OH-): \( \mathrm{S}^{2-}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{H}_{2}\mathrm{S}(aq) + 2\mathrm{OH}^{-}(aq) \). Because S2- ions react with water to produce OH- ions, the solution will become more basic.
03

Determining the pH

As sulfide ions react with water to produce OH- ions, these excess OH- ions will cause the solution to be basic. Therefore, the pH of the solution is expected to be greater than 7.

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

The Henderson-Hasselbalch equation can be written as \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}-\log \left(\frac{1}{\alpha}-1\right)\) where \(\alpha=\frac{\left[\mathrm{A}^{-}\right]}{\left[\mathrm{A}^{-}\right]+[\mathrm{HA}]}\) Thus, the degree of ionization \((\alpha)\) of an acid can be determined if both the \(\mathrm{pH}\) of the solution and the \(\mathrm{p} K_{\mathrm{a}}\) of the acid are known. (a) Use this equation to plot the pH versus the degree of ionization for the second ionization constant of phosphoric acid \(\left(K_{\mathrm{a}}=6.3 \times 10^{-8}\right)\) (b) If \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) what is the degree of ionization? (c) If the solution had a pH of 6.0 what would the value of \(\alpha\) be?

Consider a solution containing two weak monoprotic acids with dissociation constants \(K_{\mathrm{HA}}\) and \(K_{\mathrm{HB}}\). Find the charge balance equation for this system, and use it to derive an expression that gives the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) as a function of the concentrations of \(\mathrm{HA}\) and HB and the various constants.

Calculate the \(\mathrm{pH}\) of a buffer that is (a) \(0.012 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(K_{\mathrm{a}}=6.3 \times 10^{-5}\right)\) and 0.033 \(\mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) (b) \(0.408 \mathrm{M} \mathrm{NH}_{3}\) and \(0.153 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\)

Sulfuric acid is a diprotic acid, strong in the first ionization step and weak in the second \(\left(K_{\mathrm{a}_{2}}=1.1 \times 10^{-2}\right)\) By using appropriate calculations, determine whether it is feasible to titrate \(10.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to two distinct equivalence points with \(0.100 \mathrm{M} \mathrm{NaOH}\)

Sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4},\) is made commercially by first neutralizing phosphoric acid with sodium carbonate to obtain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\). The \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) is further neutralized to \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) with \(\mathrm{NaOH}\) (a) Write net ionic equations for these reactions. (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is a much cheaper base than is \(\mathrm{NaOH}\) Why do you suppose that \(\mathrm{NaOH}\) must be used as well as \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to produce \(\mathrm{Na}_{3} \mathrm{PO}_{4} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

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