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Chapter 5: Problem 31

Which solutions would you use to precipitate \(\mathrm{Mg}^{2+}\) from an aqueous solution of \(\mathrm{MgCl}_{2} ?\) Explain your choice. (a) \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (b) \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq});\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(\mathrm{aq})\).

Short Answer

Expert verified
Among the given options, only \(\mathrm{NH}_{3}\) would precipitate \(\mathrm{Mg}^{2+}\) from the aqueous solution of \(\mathrm{MgCl}_{2}\)

Step by step solution

01

Analyze solubilities

Certain salts have certain solubility rules. Magnesium salts are generally soluble, so a solution that responds with a magnesium ion to make an insoluble compound is required.
02

Analyze reactant (a): \(\mathrm{KNO}_{3}\)

Firstly look at \(\mathrm{KNO}_{3}\). The nitrate (\(\mathrm{NO}_{3}^{-}\)) ion is a part of a group of common ions that do not react to form precipitates. Therefore, \(\mathrm{KNO}_{3}\) would not precipitate \(\mathrm{Mg}^{2+}\).
03

Analyze reactant (b): \(\mathrm{NH}_{3}\)

Next, consider \(\mathrm{NH}_{3}\). Ammonia in water acts as a base and forms ammonium ion (\(\mathrm{NH}_{4}^{+}\)) and hydroxide ion (\(\mathrm{OH}^{-}\)). The hydroxide ion can react with \(\mathrm{Mg}^{2+}\) to form \(\mathrm{Mg(OH)}_{2}\), which is insoluble in water. Therefore, \(\mathrm{NH}_{3}\) would precipitate \(\mathrm{Mg}^{2+}\).
04

Analyze reactant (c): \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Next is \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The sulfate (\(\mathrm{SO}_{4}^{2-}\)) ion will react with \(\mathrm{Mg}^{2+}\) to form \(\mathrm{MgSO}_{4}\) which is soluble in water according to solubility rules, hence will not precipitate out of solution. Therefore, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) would not precipitate \(\mathrm{Mg}^{2+}\).
05

Analyze reactant (d): \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (acetic acid) is a weak acid and does not have a significant effect on magnesium ion in water, as it does not provide any ion that can form a precipitate with \(\mathrm{Mg}^{2+}\). Hence \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) would not precipitate \(\mathrm{Mg}^{2+}\).

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Most popular questions from this chapter

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) $\mathrm{Ca}^{2+}+2 \mathrm{I}^{-}+2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \longrightarrow$ (b) $\mathrm{Ba}^{2+}+\mathrm{S}^{2-}+2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow$

A piece of marble (assume it is pure \(\mathrm{CaCO}_{3}\) ) reacts with \(2.00 \mathrm{L}\) of \(2.52 \mathrm{M} \mathrm{HCl}\). After dissolution of the marble, a \(10.00 \mathrm{mL}\) sample of the resulting solution is withdrawn, added to some water, and titrated with 24.87 mL of 0.9987 M NaOH. What must have been the mass of the piece of marble? Comment on the precision of this method; that is, how many significant figures are justified in the result?

How many milliliters of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) are required to titrate \(50.00 \mathrm{mL}\) of \(0.0526 \mathrm{M} \mathrm{HNO}_{3} ?\)

Determine which of the following react(s) with HCl(ag) to produce a gas, and write a net ionic equation(s) for the reaction(s). (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4} ;\) (b) \(\mathrm{KHSO}_{3}\); (c) \(\mathrm{Zn}(\mathrm{OH})_{2};\) (d) \(\mathrm{CaCl}_{2}\).
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