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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. \( \mathrm{Fe(NO_{3})_{3}(aq) + 3NaOH(aq) \to Fe(OH)_{3}(s) + 3NaNO_{3}(aq)} \) b. \( \mathrm{Hg_{2}(NO_{3})_{2}(aq) + 2NaCl(aq) \to Hg_{2}Cl_{2}(s) + 2NaNO_{3}(aq)} \) c. \( \mathrm{Pb(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \to PbSO_{4}(s) + 2NaNO_{3}(aq)} \) d. \( \mathrm{Ba(NO_{3})_{2}(aq) + K_{2}CrO_{4}(aq) \to BaCrO_{4}(s) + 2KNO_{3}(aq)} \)

Step by step solution

01

a. Production of \(\mathrm{Fe}(\mathrm{OH})_{3}\)(s) by precipitation reaction

In order to produce \(\mathrm{Fe}(\mathrm{OH})_{3}\)(s), we will need an aqueous solution containing Fe(III) ions and another containing hydroxide ions. We can choose, for example, \(\mathrm{Fe}(\mathrm{NO}_{3})_{3}(\aq)\) as a source of Fe(III) ions and \(\mathrm{NaOH}(\aq)\) as a source of hydroxide ions. The reaction would be: \[\mathrm{Fe(NO_{3})_{3}(aq) + 3NaOH(aq) \to Fe(OH)_{3}(s) + 3NaNO_{3}(aq)}\]
02

b. Production of \(\mathrm{Hg_{2}Cl_{2}}\)(s) by precipitation reaction

In order to produce \(\mathrm{Hg_{2}Cl_{2}}\)(s), we will need an aqueous solution containing Hg(I) ions and another containing chloride ions. We can choose, for example, \(\mathrm{Hg_{2}(NO_{3})_{2}(\aq)}\) as a source of Hg(I) ions and \(\mathrm{NaCl(aq)}\) as a source of chloride ions. The reaction would be: \[\mathrm{Hg_{2}(NO_{3})_{2}(aq) + 2NaCl(aq) \to Hg_{2}Cl_{2}(s) + 2NaNO_{3}(aq)}\]
03

c. Production of \(\mathrm{PbSO_{4}}\)(s) by precipitation reaction

In order to produce \(\mathrm{PbSO_{4}}\)(s), we will need an aqueous solution containing Pb(II) ions and another containing sulfate ions. We can choose, for example, \(\mathrm{Pb(NO_{3})_{2}(aq)}\) as a source of Pb(II) ions and \(\mathrm{Na_{2}SO_{4}(aq)}\) as a source of sulfate ions. The reaction would be: \[\mathrm{Pb(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \to PbSO_{4}(s) + 2NaNO_{3}(aq)}\]
04

d. Production of \(\mathrm{BaCrO_{4}}\)(s) by precipitation reaction

In order to produce \(\mathrm{BaCrO_{4}}\)(s), we will need an aqueous solution containing Ba(II) ions and another containing chromate ions. We can choose, for example, \(\mathrm{Ba(NO_{3})_{2}(aq)}\) as a source of Ba(II) ions and \(\mathrm{K_{2}CrO_{4}(aq)}\) as a source of chromate ions. The reaction would be: \[\mathrm{Ba(NO_{3})_{2}(aq) + K_{2}CrO_{4}(aq) \to BaCrO_{4}(s) + 2KNO_{3}(aq)}\]

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Most popular questions from this chapter

Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion b. calcium ion c. iron(III) ion d. sulfate ion e. mercury(I) ion, \(\mathrm{Hg}_{2}{ }^{2+}\) f. silver ion

A solution is prepared by dissolving \(0.6706 \mathrm{~g}\) oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make \(100.0 \mathrm{~mL}\) of solution. \(\mathrm{A}\) \(10.00-\mathrm{mL}\) aliquot (portion) of this solution is then diluted to a final volume of \(250.0 \mathrm{~mL}\). What is the final molarity of the oxalic acid solution?

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The \(\mathrm{VO}^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q)\) To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.

Consider the reaction between sodium metal and fluorine \(\left(\mathrm{F}_{2}\right)\) gas to form sodium fluoride. Using oxidation states, how many electrons would each sodium atom lose, and how many electrons would each fluorine atom gain? How many sodium atoms are needed to react with one fluorine molecule? Write a balanced equation for this reaction.

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 121 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.
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