Predict in each case whether a reaction is likely to occur. If so, write a net ionic equation. (a) $\mathrm{HI}(\mathrm{aq})+\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \longrightarrow$ (b) $\operatorname{CuSO}_{4}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow$ (c) $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \longrightarrow$

Short Answer

Expert verified
(a): No reaction. \n (b): Reaction occurs. The balanced net ionic equation: Cu^{2+}(aq) + CO_{3}^{2-}(aq) \longrightarrow CuCO_{3}(s). \n (c): Reaction occurs. The balanced net ionic equation: 3Cu^{2+}(aq) + 2PO_{4}^{3-}(aq) \longrightarrow Cu_{3}(PO_{4})_{2}(s).

Step by step solution

01

Case (a): HI(aq) + Zn(NO3)2(aq)

Based on the reactivity series, hydrogen ions (H+) cannot displace zinc (Zn) from its salt. Therefore, no reaction is expected to occur.
02

Case (b): CuSO4(aq) + Na2CO3(aq)

Here, copper sulfate (CuSO4) reacts with sodium carbonate (Na2CO3) to form copper carbonate (CuCO3) and sodium sulfate (Na2SO4). Copper carbonate, being insoluble, will precipitate out of the solution. The net ionic equation for the reaction is: \[ \mathrm{Cu^{2+}(aq) + CO_{3}^{2-}(aq) \longrightarrow CuCO_{3}(s)} \]
03

Case (c): Cu(NO3)2(aq) + Na3PO4(aq)

Copper nitrate (Cu(NO3)2) reacts with sodium phosphate (Na3PO4) to produce copper phosphate (Cu3(PO4)2) which is insoluble and therefore precipitates, and sodium nitrate (NaNO3): \[ 3\mathrm{Cu^{2+}(aq) + 2PO_{4}^{3-}(aq) \longrightarrow Cu_{3}(PO_{4})_{2}(s)} \]. Please notice that the '(s)' notation after the product indicates a solid product, meaning a precipitate was formed in the reaction.

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Most popular questions from this chapter

Balance these equations for redox reactions occurring in acidic solution. (a) $\mathrm{P}_{4}(\mathrm{s})+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NO}(\mathrm{g})$ (b) $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{SO}_{4}^{2-}+\mathrm{Mn}^{2+}$ (c) $\mathrm{HS}^{-}+\mathrm{HSO}_{3}^{-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ (d) $\mathrm{Fe}^{3+}+\mathrm{NH}_{3} \mathrm{OH}^{+} \longrightarrow \mathrm{Fe}^{2+}+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})$

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) \(\mathrm{CaCl}_{2}\).

An \(\mathrm{NaOH}(\mathrm{aq})\) solution cannot be made up to an exact concentration simply by weighing out the required mass of NaOH, because the NaOH is not pure. Also, water vapor condenses on the solid as it is being weighed. The solution must be standardized by titration. For this purpose, a 25.00 \(\mathrm{mL}\) sample of an NaOH(aq) solution requires 28.34 \(\mathrm{mL}\) of 0.1085 \(\mathrm{M}\) HCl. What is the molarity of the NaOH(aq)? \(\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

A sample of battery acid is to be analyzed for its sulfuric acid content. A \(1.00 \mathrm{mL}\) sample weighs \(1.239 \mathrm{g}\). This \(1.00 \mathrm{mL}\) sample is diluted to \(250.0 \mathrm{mL}\), and \(10.00 \mathrm{mL}\) of this diluted acid requires \(32.44 \mathrm{mL}\) of \(0.00498 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. What is the mass percent of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the battery acid? (Assume that complete ionization and neutralization of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) occurs.)

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

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