Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 111

The ingestion of a very small quantity of mercury is not considered too harmful. Would this statement still hold if the gastric juice in your stomach were mostly nitric acid instead of hydrochloric acid?

Short Answer

Expert verified
No, if the gastric juice were mostly nitric acid instead of hydrochloric acid, ingestion of mercury would not be harmless as mercury reacts with nitric acid to form mercuric nitrate, which can cause mercury poisoning.

Step by step solution

01

Reaction of Mercury with Hydrochloric Acid

Mercury doesn't react with hydrochloric acid, even when heated. So, ingesting a small quantity of mercury with hydrochloric acid (which is naturally present in stomach) would not have chemical reactions that produce harmful substances.
02

Reactions of Mercury with Nitric Acid

However, Mercury is known to react with nitric acid, as shown by the equation: \( Hg_{(l)} + 8HNO3_{(aq)} \rightarrow 4Hg(NO3)2_{(aq)} + 2NO_{(g)} + 4H2O_{(l)} \). This reaction generates mercuric nitrate \(Hg(NO3)2\), nitrogen monoxide \(NO\), and water. Mercuric nitrate is soluble in water, so it could be absorbed by body tissues, causing mercury poisoning.
03

Final Conclusion

Therefore, if the gastric juice in your stomach were mostly nitric acid instead of hydrochloric acid, ingestion of a very small quantity of mercury would not be harmless. It could lead to mercury poisoning due to the formation of mercuric nitrate.

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

Show a sketch of a galvanic concentration cell. Each compartment consists of a Co electrode in a \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) solution. The concentrations in the compartments are \(2.0 \mathrm{M}\) and \(0.10 \mathrm{M}\), respectively. Label the anode and cathode compartments. Show the direction of electron flow. (a) Calculate the \(E_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\). (b) What are the concentrations in the compartments when the \(E_{\text {cell }}\) drops to 0.020 V? Assume volumes to remain constant at \(1.00 \mathrm{~L}\) in each compartment.

The half-reaction at an electrode is $$\mathrm{Mg}^{2+}(\text { molten })+2 e^{-} \longrightarrow \mathrm{Mg}(s)$$ Calculate the number of grams of magnesium that can be produced by supplying \(1.00 \mathrm{~F}\) to the electrode.

From the following information, calculate the solubility product of AgBr: \(\mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \operatorname{Ag}(s) \quad E^{\circ}=0.80 \mathrm{~V}\) \(\operatorname{AgBr}(s)+e^{-} \longrightarrow \operatorname{Ag}(s)+\operatorname{Br}^{-}(a q) \quad E^{\circ}=0.07 \mathrm{~V}\)

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(P_{\mathrm{H}}=2.0 \mathrm{~atm} ?\)

The diagram here shows an electrolytic cell consisting of a Co electrode in a \(2.0 \mathrm{M} \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) solution and a Mg electrode in a \(2.0 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) solution. (a) Label the anode and cathode and show the halfcell reactions. Also label the signs \((+\) or \(-)\) on the battery terminals. (b) What is the minimum voltage to drive the reaction? (c) After the passage of \(10.0 \mathrm{~A}\) for \(2.00 \mathrm{~h}\) the battery is replaced with a voltmeter and the electrolytic cell now becomes a galvanic cell. Calculate \(E_{\text {cell. }}\) Assume volumes to remain constant at \(1.00 \mathrm{~L}\) in each compartment.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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