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Chapter 4: Problem 8

The labels have fallen off two bottles, one containing \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and the other containing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). You have a bottle of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\). How could you use it to test a portion of each solution to identify which solution is which? [Section 4.2]

Short Answer

Expert verified
Add dilute sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) to each unknown solution. If the resulting solution is clear, the initial solution is magnesium nitrate (\(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)). If a precipitate forms, the initial solution is lead nitrate (\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)).

Step by step solution

01

Write down the chemical reactions with sulfuric acid

We will add dilute sulfuric acid to both solutions: 1) Magnesium nitrate reacts with sulfuric acid: \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{MgSO}_{4} + 2\mathrm{HNO}_{3}\) 2) Lead nitrate reacts with sulfuric acid: \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{PbSO}_{4鈫搣 + 2\mathrm{HNO}_{3}\)
02

Analyze the results of the reactions

Now, let's analyze the products of the reactions: 1) Magnesium sulfate (\(\mathrm{MgSO}_{4}\)) is a soluble salt and will dissolve in water, forming a clear solution. Nitric acid (\(2\mathrm{HNO}_{3}\)) is also a soluble compound. 2) Lead sulfate (\(\mathrm{PbSO}_{4}\)) is an insoluble salt and will form a precipitate when mixed with sulfuric acid. Nitric acid (\(2\mathrm{HNO}_{3}\)) is also a soluble compound.
03

Perform the test and identify the solutions

In order to identify which solution is which, we will add the dilute sulfuric acid to each unknown solution: 1) If the resulting solution is clear after adding sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)), then the initial solution is magnesium nitrate (\(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)). 2) If the resulting solution forms a precipitate after adding sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)), then the initial solution is lead nitrate (\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)).

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Most popular questions from this chapter

The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), which reacts with sodium hydroxide in the following fashion: \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) $$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of \(0.115 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a \(1.00\) -qt sample of this vinegar?

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: \(2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) $$ Sodium bicarbonate is added untilthe fizzing due to the formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

(a) Calculate the molarity of a solution made by dissolving \(0.750\) grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in enough water to form exactly \(850 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{KMnO}_{4}\) are present in \(250 \mathrm{~mL}\) of a \(0.0475 \mathrm{M}\) solution? (c) How many milliliters of \(11.6 \mathrm{M} \mathrm{HCl}\) solution are needed to obtain \(0.250 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

(a) Is the concentration of a solution an intensive or an extensive property? (b) What is the difference between \(0.50 \mathrm{~mol} \mathrm{HCl}\) and \(0.50 \mathrm{M} \mathrm{HCl} ?\)

When asked what causes electrolyte solutions to conduct electricity, a student responds that it is due to the movement of electrons through the solution. Is the student correct? If not, what is the correct response?
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