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Chapter 17: Problem 62

Describe a simple test that would enable you to distinguish between \(\mathrm{AgNO}_{3}(s)\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\).

Short Answer

Expert verified
To distinguish between \(\mathrm{AgNO}_{3}(s)\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\), introduce a sodium chloride solution to aqueous solutions of both. The \(\mathrm{AgNO}_{3}(s)\) solution will form a white precipitate of silver chloride while the \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\) solution will remain clear.

Step by step solution

01

Prepare solutions of the two substances

First, dissolve a small amount of the two substances in distilled water separately to get their aqueous solutions.
02

Add Sodium Chloride (NaCl) solution

Take two separate test tubes, add each solution into each of the test tubes, and then introduce a sodium chloride (NaCl) solution into both tubes.
03

Observe and compare the reactions

\(\mathrm{AgNO}_{3}(s)\) will react with the sodium chloride to form a white precipitate of silver chloride \(\mathrm{AgCl}\), whereas \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\) will not form a precipitate, keeping the solution clear.

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Most popular questions from this chapter

A sample of \(0.1276 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0633 \mathrm{M} \mathrm{NaOH}\) solution. The volume of base required to reach the equivalence point was \(18.4 \mathrm{~mL}\). (a) Calculate the molar mass of the acid. (b) After \(10.0 \mathrm{~mL}\) of base had been added to the titration, the \(\mathrm{pH}\) was determined to be 5.87 . What is the \(K_{\mathrm{a}}\) of the unknown acid?

Give two examples of metal ions in each group (1 through 5) in the qualitative analysis scheme.

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A sample of \(20.0 \mathrm{~mL}\) of \(0.10 \mathrm{MBa}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\). Will \(\mathrm{BaCO}_{3}\) precipitate?

A quantity of \(0.560 \mathrm{~g}\) of \(\mathrm{KOH}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). Excess \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is then added to the solution. What mass (in grams) of \(\mathrm{CO}_{2}\) is formed?
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