Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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) Net ionic: \( 2 \text{H}^+(aq) + \text{2 OH}^-(aq) \rightarrow 2 \text{H}_2O(l) \); (b) Net ionic: \( \text{Cu(OH)}_2(s) + 2 \text{H}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2 \text{H}_2O(l) \); (c) Net ionic: \( \text{Al(OH)}_3(s) + 3 \text{H}^+(aq) \rightarrow \text{Al}^{3+}(aq) + 3 \text{H}_2O(l) \).

Step by step solution

01

Write Complete Molecular Equation for (a)

Identify and combine the reactants \( \text{HBr}(aq) \) and \( \text{Ca(OH)}_2(aq) \) to form products. The expected products will be \( \text{CaBr}_2(aq) \) and \( \text{H}_2O(l) \). Thus, the complete molecular equation is: \[ \text{2 HBr}(aq) + \text{Ca(OH)}_2(aq) \rightarrow \text{CaBr}_2(aq) + 2 \text{H}_2O(l) \]
02

Write Net Ionic Equation for (a)

Identify and remove spectator ions. In this reaction, \( \text{Br}^- \) and \( \text{Ca}^{2+} \) are spectator ions. The net ionic equation involves the following: \[ 2 \text{H}^+(aq) + \text{2 OH}^-(aq) \rightarrow 2 \text{H}_2O(l) \]
03

Write Complete Molecular Equation for (b)

Combine the reactants \( \text{Cu(OH)}_2(s) \) and \( \text{HClO}_4(aq) \). The expected products are \( \text{Cu(ClO}_4)_2(aq) \) and \( \text{H}_2O(l) \): \[ \text{Cu(OH)}_2(s) + 2 \text{HClO}_4(aq) \rightarrow \text{Cu(ClO}_4)_2(aq) + 2 \text{H}_2O(l) \]
04

Write Net Ionic Equation for (b)

Since \( \text{Cu(OH)}_2(s) \) does not dissociate, the net ionic equation for this reaction is: \[ \text{Cu(OH)}_2(s) + 2 \text{H}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2 \text{H}_2O(l) \]
05

Write Complete Molecular Equation for (c)

For the reaction \( \text{Al(OH)}_3(s) \) with \( \text{HNO}_3(aq) \), the products formed are \( \text{Al(NO}_3)_3(aq) \) and \( \text{H}_2O(l) \). The balanced molecular equation is: \[ \text{Al(OH)}_3(s) + 3 \text{HNO}_3(aq) \rightarrow \text{Al(NO}_3)_3(aq) + 3 \text{H}_2O(l) \]
06

Write Net Ionic Equation for (c)

With \( \text{Al(OH)}_3(s) \) not dissociating in solution, the net ionic equation is: \[ \text{Al(OH)}_3(s) + 3 \text{H}^+(aq) \rightarrow \text{Al}^{3+}(aq) + 3 \text{H}_2O(l) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a crucial skill to ensure that the number of atoms for each element is the same on both sides of the equation. This process respects the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. When balancing equations:
  • First, write the correct formulas for all reactants and products.
  • Count the number of atoms of each element present in the reactants and products.
  • Adjust coefficients—the numbers in front of the formulas—to get the same number of each type of atom on both sides.
  • Start with the most complex substance, adjusting coefficients, not subscripts.
  • Finally, check to ensure that all atoms are balanced and that coefficients are in the smallest possible ratio.
For example, in the equation between hydrogen bromide and calcium hydroxide, the balanced equation ensures that both hydrogen and oxygen are accounted for on both sides ensuring no atoms are lost in the process.
Spectator Ions
Spectator ions are ions that appear in the same form on both the reactant and product sides of a chemical equation. They do not participate in the actual chemical change but are present to balance the charge in solution. Identifying and removing spectator ions is an essential step to derive the net ionic equation. These ions are often the ones that remain dissolved and unchanged as the reaction takes place, commonly seen in aqueous solutions. For instance, in a typical reaction with acids and bases:
  • The ions that make up soluble products remain as spectator ions.
  • They help in maintaining electrical neutrality in the solution.
  • When forming net ionic equations, spectators are left out to highlight the actual chemical change.
Recognizing spectator ions helps simplify complex reactions and focus on the core changes happening chemically.
Chemical Reactions
Chemical reactions involve the process where reactants transform into products, showcasing new substances with different properties. There are various types of reactions, including synthesis, decomposition, single-replacement, double-replacement, and combustion. Key points about chemical reactions:
  • They involve breaking existing bonds and forming new ones to create new materials.
  • Conservation of energy applies; the energy required to break bonds is offset by the energy released when new bonds form.
  • Indicators such as temperature change, gas formation, and color change can suggest a chemical reaction has occurred.
When writing net ionic equations for chemical reactions, it’s essential to focus on the actual chemical changes. For example, in the reaction between aluminum hydroxide and nitric acid, the transformation involves creating soluble aluminum ions and water, illustrating the reaction’s core event. This understanding lays the foundation for predicting products and understanding overall reaction mechanics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In 2014 , a major chemical leak at a facility in West Virginia released \(28,390 \mathrm{~L}\) of MCHM (4-methylcyclohexylmethanol, \(\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}\) ) into the Elk River. The density of MCHM is 0.9074 \(\mathrm{g} / \mathrm{mL}\). (a) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is \(2.00 \mathrm{~m}\) deep, \(90.0 \mathrm{~m}\) wide, and \(90.0 \mathrm{~m}\) long. (b) How much farther down the river would the spill have to spread in order to achieve a "safe" MCHM concentration of \(1.00 \times 10^{-4} \mathrm{M}\) ? Assume the depth and width of the river are constant and the concentration of MCHM is uniform along the length of the spill.

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}^{3-},\) what mass of sodium arsenate would be present in a \(1.00-\mathrm{L}\) sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9} $$

A person suffering from hyponatremia has a sodium ion concentration in the blood of \(0.118 \mathrm{M}\) and a total blood volume of \(4.6 \mathrm{~L} .\) What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to \(0.138 M\), assuming no change in blood volume?

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 resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+& \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ 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?

Calculate the concentration of each ion in the following solutions obtained by mixing: (a) \(32.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{KMnO}_{4}\) with \(15.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{KMnO}_{4}\) (b) \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{ZnCl}_{2}\) with \(5.0 \mathrm{~mL}\) of \(0.200 \mathrm{MZn}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) 4.2 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) in \(150.0 \mathrm{~mL}\) of \(0.02 M \mathrm{KCl}\) solution. Assume that the volumes are additive.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free