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Chapter 20: Problem 45

A solution of potassium dichromate is made basic with sodium hydroxide; the color changes from red to yellow. Addition of silver nitrate to the yellow solution gives a precipitate. This precipitate dissolves in concentrated ammonia but re-forms when nitric acid is added. Write balanced net ionic equations for all the reactions in this sequence.

Short Answer

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Question: Write down the balanced net ionic equations for each step of the following reactions: 1) Potassium dichromate solution turning basic with sodium hydroxide; 2) Adding silver nitrate to the yellow solution to form a precipitate; 3) Dissolving the precipitate in concentrated ammonia; 4) Re-forming the precipitate by adding nitric acid. Answer: 1) Cr2O7^2- + 2OH^- → 2CrO4^2- 2) 2Ag^+ + CrO4^2- → Ag2CrO4(s) 3) Ag2CrO4(s) + 4NH3 → 2Ag(NH3)2^+ + CrO4^2- 4) 2Ag(NH3)2^+ + CrO4^2- + 4H^+ → Ag2CrO4(s) + 4NH4^+

Step by step solution

01

Write the chemical formulas of the compounds involved in the reactions

We will determine the chemical formulas of each compound mentioned in the exercise. Potassium dichromate: K2Cr2O7 Sodium hydroxide: NaOH Silver nitrate: AgNO3 Ammonia: NH3 Nitric acid: HNO3 Now we can analyze the reactions and write the balanced net ionic equations.
02

Formation of basic potassium dichromate solution

In this step, potassium dichromate (K2Cr2O7) reacts with sodium hydroxide (NaOH) to form a basic solution with a yellow color. The balanced chemical equation for this reaction is: K2Cr2O7 + 2NaOH → 2KOH + Na2CrO4 The net ionic equation is obtained by eliminating the spectator ions, which are potassium (K+) and sodium (Na+) ions, as they remain unchanged during the reaction. Net ionic equation: Cr2O7^2- + 2OH^- → 2CrO4^2-
03

Formation of a precipitate by adding silver nitrate

In this reaction, silver nitrate (AgNO3) is added to the yellow solution from step 2. The silver ions (Ag+) react with the chromate ions (CrO4^2-) to form a precipitate, silver chromate (Ag2CrO4). AgNO3 + Na2CrO4 → Ag2CrO4(s) + 2NaNO3 Cancel out the spectator ions (Na+ and NO3^-): Net ionic equation: 2Ag^+ + CrO4^2- → Ag2CrO4(s)
04

Precipitate dissolving in concentrated ammonia

Silver chromate (Ag2CrO4) precipitate will dissolve in ammonia (NH3) to form a complex ion. Ag2CrO4(s) + 4NH3 → 2Ag(NH3)2^+ + CrO4^2- Since there are no spectator ions, the net ionic equation is the same as the balanced chemical equation. Net ionic equation: Ag2CrO4(s) + 4NH3 → 2Ag(NH3)2^+ + CrO4^2-
05

Precipitate re-forming when nitric acid is added

In this final step, nitric acid (HNO3) is added to the solution, causing the silver complex ion to decompose and reform the silver chromate (Ag2CrO4) precipitate. 2Ag(NH3)2^+ + CrO4^2- + 4HNO3 → Ag2CrO4(s) + 4NH4NO3 Eliminate the spectator ions (NO3^-): Net ionic equation: 2Ag(NH3)2^+ + CrO4^2- + 4H^+ → Ag2CrO4(s) + 4NH4^+ These are the balanced net ionic equations for each reaction in this sequence.

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Most popular questions from this chapter

Write a balanced equation to show (a) the formation of gas bubbles when cobalt reacts with hydrochloric acid. (b) the reaction of copper with nitric acid. (c) the reduction half-reaction of dichromate ion to \(\mathrm{Cr}^{3+}\) in acid solution.

Write a balanced equation to represent the roasting of copper(I) sulfide to form "blister copper"

A \(0.500-g\) sample of steel is analyzed for manganese. The sample is dissolved in acid and the manganese is oxidized to permanganate ion. A measured excess of \(\mathrm{Fe}^{2+}\) is added to reduce \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\). The excess \(\mathrm{Fe}^{2+}\) is determined by titration with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{27}\). If \(75.00 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{FeSO}_{4}\) is added and the excess requires \(13.50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to oxidize \(\mathrm{Fe}^{2+}\), calculate the percent by mass of \(\mathrm{Mn}\) in the sample.

A \(0.500-\mathrm{g}\) sample of zinc-copper alloy was treated with dilute hydrochloric acid. The hydrogen gas evolved was collected by water displacement at \(27^{\circ} \mathrm{C}\) and a total pressure of \(755 \mathrm{~mm} \mathrm{Hg}\). The volume of the water displaced by the gas is \(105.7 \mathrm{~mL}\). What is the percent composition, by mass, of the alloy? (Vapor pressure of \(\mathrm{H}_{2} \mathrm{O}\) at \(27^{\circ} \mathrm{C}\) is \(26.74 \mathrm{~mm} \mathrm{Hg}\).) Assume only the zinc reacts.

Give the formula and name of the compound formed by strontium with (a) nitrogen (b) bromine (c) water (d) oxygen
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