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Chapter 4: Problem 50

Give an example how each of the following insoluble ionic compounds could be produced using a precipitation reaction. Write the balanced formula equation for each reaction. a. \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\) c. \(\mathrm{PbSO}_{4}(s)\) d. \(\mathrm{BaCrO}_{4}(s)\)

Short Answer

Expert verified
a. Iron (III) hydroxide: \[2\mathrm{Fe(NO}_3)_3(aq) + 6\mathrm{NaOH}(aq) \rightarrow 2\mathrm{Fe(OH)}_3(s) + 6\mathrm{NaNO_3}(aq)\] b. Mercury (I) chloride: \[\mathrm{Hg_2(NO}_3)_2(aq) + 2\mathrm{NaCl}(aq) \rightarrow \mathrm{Hg_2Cl_2}(s) + 2\mathrm{NaNO_3}(aq)\] c. Lead (II) sulfate: \[\mathrm{Pb(NO}_3)_2(aq) + \mathrm{Na_2SO}_4(aq) \rightarrow \mathrm{PbSO}_4(s) + 2\mathrm{NaNO_3}(aq)\] d. Barium chromate: \[\mathrm{Ba(NO}_3)_2(aq) + \mathrm{K_2CrO}_4(aq) \rightarrow \mathrm{BaCrO}_4(s) + 2\mathrm{KNO_3}(aq)\]

Step by step solution

01

a. Formation of Iron (III) Hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\)

To form \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\), we can mix a soluble iron (III) compound, such as iron (III) nitrate, \(\mathrm{Fe(NO}_3)_3(aq)\), and a soluble hydroxide compound, such as sodium hydroxide, \(\mathrm{NaOH}(aq)\). The balanced equation for this precipitation reaction is: \[2\mathrm{Fe(NO}_3)_3(aq) + 6\mathrm{NaOH}(aq) \rightarrow 2\mathrm{Fe(OH)}_3(s) + 6\mathrm{NaNO_3}(aq)\]
02

b. Formation of Mercury (I) Chloride, \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\)

To form \(\mathrm{Hg}_{2}\mathrm{Cl}_{2}(s)\), we can mix a soluble mercury (I) compound, such as mercury (I) nitrate, \(\mathrm{Hg_2(NO}_3)_2(aq)\), and a soluble chloride compound, such as sodium chloride, \(\mathrm{NaCl}(aq)\). The balanced equation for this precipitation reaction is: \[\mathrm{Hg_2(NO}_3)_2(aq) + 2\mathrm{NaCl}(aq) \rightarrow \mathrm{Hg_2Cl_2}(s) + 2\mathrm{NaNO_3}(aq)\]
03

c. Formation of Lead (II) Sulfate, \(\mathrm{PbSO}_{4}(s)\)

To form \(\mathrm{PbSO}_4(s)\), we can mix a soluble lead (II) compound, such as lead (II) nitrate, \(\mathrm{Pb(NO}_3)_2(aq)\), and a soluble sulfate compound, such as sodium sulfate, \(\mathrm{Na_2SO}_4(aq)\). The balanced equation for this precipitation reaction is: \[\mathrm{Pb(NO}_3)_2(aq) + \mathrm{Na_2SO}_4(aq) \rightarrow \mathrm{PbSO}_4(s) + 2\mathrm{NaNO_3}(aq)\]
04

d. Formation of Barium Chromate, \(\mathrm{BaCrO}_{4}(s)\)

To form \(\mathrm{BaCrO}_4(s)\), we can mix a soluble barium compound, such as barium nitrate, \(\mathrm{Ba(NO}_3)_2(aq)\), and a soluble chromate compound, such as potassium chromate, \(\mathrm{K_2CrO}_4(aq)\). The balanced equation for this precipitation reaction is: \[\mathrm{Ba(NO}_3)_2(aq) + \mathrm{K_2CrO}_4(aq) \rightarrow \mathrm{BaCrO}_4(s) + 2\mathrm{KNO_3}(aq)\]

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Most popular questions from this chapter

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3} \quad\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) d. As i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) e. \(\mathrm{HAsO}_{2}\)

A \(10.00-\mathrm{mL}\) sample of vinegar, an aqueous solution of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\), is titrated with \(0.5062 \mathrm{M} \mathrm{NaOH}\), and \(16.58 \mathrm{~mL}\) is required to reach the equivalence point. a. What is the molarity of the acetic acid? b. If the density of the vinegar is \(1.006 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass percent of acetic acid in the vinegar?

The soda you are drinking contains \(0.5 \%\) by mass sodium benzoate as a preservative. What approximate mass of sodium benzoate is contained in \(1.00 \mathrm{~L}\) of the solution assuming that the density of the soda is \(1.00 \mathrm{~g} / \mathrm{mL}\) (the approximate density of water)?

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 \mathrm{M} \mathrm{NaOH}\) from \(1.00 \mathrm{M} \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

Calculate the molarity of each of these solutions. a. A \(5.623-\mathrm{g}\) sample of \(\mathrm{NaHCO}_{3}\) is dissolved in enough water to make \(250.0 \mathrm{~mL}\) of solution. b. A \(184.6-\mathrm{mg}\) sample of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is dissolved in enough water to make \(500.0 \mathrm{~mL}\) of solution. c. A 0.1025-g sample of copper metal is dissolved in \(35 \mathrm{~mL}\) of concentrated \(\mathrm{HNO}_{3}\) to form \(\mathrm{Cu}^{2+}\) ions and then water is added to make a total volume of \(200.0 \mathrm{~mL}\). (Calculate the molarity of \(\mathrm{Cu}^{2+} .\) )
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