/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Give an example how each of the ... [FREE SOLUTION] | 91影视

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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

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584 \mathrm{~g}\) pure manganese metal in nitric acid and diluting to a final volume of \(1.000 \mathrm{~L}\). The following solutions were then prepared by dilution: For solution A. \(50.00 \mathrm{~mL}\) of stock solution was diluted to \(1000.0 \mathrm{~mL}\) For solution \(B, 10.00 \mathrm{~mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{~mL}\). For solution \(C, 10.00 \mathrm{~mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{~mL}\). Calculate the concentrations of the stock solution and solutions \(A, B\), and \(C\).

A \(30.0\) -mL sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{~mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is \(0.0300 M\), give some possible identities for the unknown base.

You made \(100.0 \mathrm{~mL}\) of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only \(80.0 \mathrm{~mL}\) left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take \(2.00 \mathrm{~mL}\) of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of \(3.407 \mathrm{~g}\). What was the concentration of the original lead(II) nitrate solution?

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50 M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 M)\) sulfuric acid b. \(\mathrm{HCl}\) from "concentrated" \((12 \mathrm{M})\) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?
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