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Chapter 4: Problem 39

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Short Answer

Expert verified
(a) Balanced molecular equation: HBr + Ca(OH)₂ → 2 H₂O + CaBr₂ \\ Net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l) (b) Balanced molecular equation: Cu(OH)₂(s) + 2 HClO₄(aq) → 2 H₂O(l) + Cu(ClO₄)₂(aq) \\ Net ionic equation: Cu(OH)₂(s) + 2 H⁺(aq) → Cu²⁺(aq) + 2 H₂O(l) (c) Balanced molecular equation: Al(OH)₃(s) + 3 HNO₃(aq) → 3 H₂O(l) + Al(NO₃)₃(aq) \\ Net ionic equation: Al(OH)₃(s) + 3 H⁺(aq) → Al³⁺(aq) + 3 H₂O(l)

Step by step solution

01

(a) Completing and balancing the molecular equation for HBr and Ca(OH)₂

First, notice that HBr (hydrobromic acid) and Ca(OH)₂ (calcium hydroxide) are both aqueous solutions. The products formed in this reaction will be water (H₂O) and a soluble salt. Since HBr is an acid and Ca(OH)₂ is a base, this reaction is an acid-base neutralization reaction. The products formed are water and calcium bromide (CaBr₂): HBr + Ca(OH)₂ → H₂O + CaBr₂ To balance the equation, we see that there are 2 bromine atoms and 2 hydroxide ions on the reactants' side. Since there are 2 bromine atoms in CaBr₂, we only need to multiply H₂O by 2 to balance the hydrogen in OH and HBr, So the balanced equation is: HBr + Ca(OH)₂ → 2 H₂O + CaBr₂
02

(a) Writing the total and net ionic equation for HBr and Ca(OH)₂

Now we need to write the total ionic equation by breaking up the soluble substances into their individual ions in the solution: H⁺(aq) + Br⁻(aq) + Ca²⁺(aq) + 2 OH⁻(aq) → 2 H₂O(l) + Ca²⁺(aq) + 2 Br⁻(aq) By examining the equation, we can identify the spectator ions that remain unchanged, which are Ca²⁺ and Br⁻. We can then remove them from the equation, yielding the net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)
03

(b) Completing and balancing the molecular equation for Cu(OH)₂ and HClO₄

We can see that Cu(OH)₂ is a solid and the reaction with HClO₄ (perchloric acid) is another acid-base reaction. The products of this reaction are water and copper(II) perchlorate salt (Cu(ClO₄)₂): Cu(OH)₂(s) + HClO₄(aq) → H₂O(l) + Cu(ClO₄)₂(aq) To balance the equation, we need 2 HClO₄ molecules to balance the 2 OH⁻ ions in Cu(OH)₂: Cu(OH)₂(s) + 2 HClO₄(aq) → 2 H₂O(l) + Cu(ClO₄)₂(aq)
04

(b) Writing the total and net ionic equation for Cu(OH)₂ and HClO₄

Now we write the total ionic equation: Cu(OH)₂(s) + 2 H⁺(aq) + 2 ClO₄⁻(aq) → 2 H₂O(l) + Cu²⁺(aq) + 2 ClO₄⁻(aq) The spectator ions are 2 ClO₄⁻ ions. The net ionic equation will be: Cu(OH)₂(s) + 2 H⁺(aq) → Cu²⁺(aq) + 2 H₂O(l)
05

(c) Completing and balancing the molecular equation for Al(OH)₃ and HNO₃

In this reaction, Al(OH)₃ is a solid and reacts with HNO₃ (nitric acid) in another acid-base reaction. This reaction produces water and aluminum nitrate salt (Al(NO₃)₃): Al(OH)₃(s) + HNO₃(aq) → H₂O(l) + Al(NO₃)₃(aq) We have 3 OH⁻ in Al(OH)₃(s) and that requires 3 HNO₃ molecules: Al(OH)₃(s) + 3 HNO₃(aq) → 3 H₂O(l) + Al(NO₃)₃(aq)
06

(c) Writing the total and net ionic equation for Al(OH)₃ and HNO₃

The total ionic equation for this reaction is: Al(OH)₃(s) + 3 H⁺(aq) + 3 NO₃⁻(aq) → 3 H₂O(l) + Al³⁺(aq) + 3 NO₃⁻(aq) The spectator ions are 3 NO₃⁻ ions. The net ionic equation is: Al(OH)₃(s) + 3 H⁺(aq) → Al³⁺(aq) + 3 H₂O(l)

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Most popular questions from this chapter

Calicheamicin gamma-1, $\mathrm{C}_{\mathrm{ss}} \mathrm{H}_{74} \mathrm{IN}_{3} \mathrm{O}_{21} \mathrm{~S}_{4},$ is one of the most potent antibiotics known: one molecule kills one bacterial cell. Describe how you would (carefully!) prepare \(25.00 \mathrm{~mL}\) of an aqueous calicheamicin gamma- 1 solution that could kill \(1.0 \times 10^{8}\) bacteria, starting from a \(5.00 \times 10^{-9} \mathrm{M}\) stock solution of the antibiotic.

Which ions remain in solution, unreacted, after each of the following pairs of solutions is mixed? (a) potassium carbonate and magnesium sulfate (b) lead nitrate and lithium sulfide (c) ammonium phosphate and calcium chloride

You want to analyze a silver nitrate solution. What mass of \(\mathrm{NaCl}\) is needed to precipitate \(\mathrm{Ag}^{+}\) ions from \(45.0 \mathrm{~mL}\) of \(0.2500 \mathrm{MAgNO}_{3}\) solution?

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2},\) or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} . \mathrm{A}\) friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) Silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a $42.3 \mathrm{~mL}$ aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?
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