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Three solutions are mixed together to form a single solution. One contains \(0.2 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2},\) the second contains \(0.1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S},\) and the third contains \(0.1 \mathrm{~mol} \mathrm{CaCl}_{2}\). (a) Write the net ionic equations for the precipitation reaction or reactions that occur. (b) What are the spectator ions in the solution?

Short Answer

Expert verified
a. The net ionic equation for the precipitation reaction is: \(Pb^{2+}(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s)\) b. The spectator ions are \(Na^+\), \(S^{2-}\), \(Ca^{2+}\), and \(CH_3COO^-\).

Step by step solution

01

Identify possible precipitation reactions

Based on the solubility rules: - All sodium salts are soluble, so no precipitate will be formed with Na鈧係. - Most acetates are soluble, so no precipitate will be formed with Pb(CH鈧僀OO)鈧 - Most chlorides are soluble, but exceptions include silver, lead, and mercury. CaCl鈧 might form a precipitate with Pb(CH鈧僀OO)鈧. Since Pb(CH鈧僀OO)鈧 and CaCl鈧 may form an insoluble compound, we'll investigate the possible precipitation reaction between them.
02

Write the balanced equation for the possible reaction

The balanced equation for the reaction between Pb(CH鈧僀OO)鈧 and CaCl鈧 is: Pb(CH鈧僀OO)鈧(aq) + CaCl鈧(aq) -> PbCl鈧(s) + Ca(CH鈧僀OO)鈧(aq)
03

Identify the insoluble compound

According to the solubility rules, PbCl鈧 is insoluble. Thus, the precipitation reaction occurs as follows: Pb(CH鈧僀OO)鈧(aq) + CaCl鈧(aq) -> PbCl鈧(s) + Ca(CH鈧僀OO)鈧(aq)
04

Write the net ionic equation for the precipitation reaction

The net ionic equation includes only the ions and compounds that participate in the precipitation reaction. In this case, it is: Pb虏鈦(aq) + 2Cl鈦(aq) -> PbCl鈧(s)
05

Identify the spectator ions

Spectator ions are the ions that do not participate in the precipitation reaction. In this case: - Na鈦 and S虏鈦 from Na鈧係 - Ca虏鈦 and CH鈧僀OO鈦 from Ca(CH鈧僀OO)鈧 So, the spectator ions are Na鈦, S虏鈦, Ca虏鈦, and CH鈧僀OO鈦. a. The net ionic equation for the precipitation reaction is: Pb虏鈦(aq) + 2Cl鈦(aq) -> PbCl鈧(s) b. The spectator ions are Na鈦, S虏鈦, Ca虏鈦, and CH鈧僀OO鈦.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reaction
In chemistry, a precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid known as a precipitate. These reactions often involve the exchange of ions between reactive species that combine to form a new compound. For example, when you mix a solution containing lead acetate, Pb(CH鈧僀OO)鈧, with a solution of calcium chloride, CaCl鈧, one product that may form is lead chloride, PbCl鈧, which is a precipitate.
The formation of a precipitate is a clear indication of a chemical reaction since a solid forms out of a liquid solution. This particular process helps in identifying the formation of new products in a reaction. When writing equations for these reactions, it鈥檚 important to focus on the substances that change states. Only these are part of the net ionic equation, as they show the actual chemical change taking place.
In the given problem, lead ions, Pb虏鈦, from lead acetate and chloride ions, Cl鈦, from calcium chloride will combine to produce lead chloride, PbCl鈧, the solid precipitate. Exploring precipitation reactions is crucial for understanding many natural and industrial processes, such as water treatment and mineral formation.
Spectator Ions
During a chemical reaction, not all ions participate in forming a new compound. Some ions remain unchanged in the solution, playing no direct role in the precipitation reaction. These are called spectator ions. They are present in the same form on both the reactant and product sides of a chemical equation.
When we write net ionic equations, we focus entirely on the ions that actively participate in forming the precipitate. Thus, spectator ions are omitted from these equations. Their presence is still necessary, as they maintain the charge balance in the solution.
In our example, the spectator ions are those that do not participate in forming lead chloride. They include Na鈦 and S虏鈦 from sodium sulfide, and Ca虏鈦 and CH鈧僀OO鈦 from calcium acetate formed by the dissolution of calcium chloride and lead acetate in water. Identifying these ions is crucial for simplifying the equations and clearly understanding the main reactions occurring in the solution.
Solubility Rules
Solubility rules provide guidelines on determining whether a compound is soluble or insoluble in water. These rules are essential when predicting the formation of a precipitate during a reaction. Typically, some ion-exchange reactions result in the formation of an insoluble substance, prompting a precipitation reaction.
  • Most sodium, potassium, and ammonium salts are soluble.
  • All nitrates and acetates are soluble.
  • Chlorides are generally soluble, with exceptions such as silver and lead chlorides.

Using these solubility guidelines, you can determine potential products in a given solution. In the example problem, while sodium or acetate salts are usually soluble, lead chloride is not, highlighting why it forms a precipitate. Understanding solubility rules allows chemists to predict whether a particular compound formed from ionic interactions will remain in solution or precipitate out as a solid.

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Most popular questions from this chapter

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

(a) Calculate the molarity of a solution made by dissolving 12.5 grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly \(550 \mathrm{~mL}\) of solution. (b) How many moles of KBr are present in \(150 \mathrm{~mL}\) of a \(0.275 \mathrm{M}\) solution? (c) How many milliliters of 6. \(1 \mathrm{M}\) HCl solution are needed to obtain \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl} ?\)

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: $$ \begin{array}{r} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{array} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathrm{b}) \mathrm{NO}_{3}^{-},(\mathrm{c}) \mathrm{NH}_{4}^{+},(\mathrm{d}) \mathrm{S}^{2-},\) (e) \(\mathrm{SO}_{4}^{2-}\). Explain briefly. [Section \(\left.4.2\right]\)

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\)

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