/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 95 Antacids are often used to relie... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 95

Antacids are often used to relieve pain and promote healing in the treatment of mild ulcers. Write balanced net ionic equations for the reactions between the HCl(aq) in the stomach and each of the following substances used in various antacids: (a) \(\mathrm{Al}(\mathrm{OH})_{3}(s)\), (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\) (c) \(\mathrm{MgCO}_{3}(s)\) (d) \(\mathrm{NaAl}\left(\mathrm{CO}_{3}\right)(\mathrm{OH})_{2}(s),(\mathrm{e}) \mathrm{CaCO}_{3}(s)\)

Short Answer

Expert verified
The net ionic equations for the reaction of HCl(aq) with the given antacids are: (a) \(\mathrm{Al(OH)_{3}(s)} + 3\mathrm{H^+(aq)} \rightarrow \mathrm{Al^{3+}(aq)} + 3\mathrm{H_{2}O(l)}\) (b) \(\mathrm{Mg(OH)_{2}(s)} + 2\mathrm{H^+(aq)} \rightarrow \mathrm{Mg^{2+}(aq)} + 2\mathrm{H_{2}O(l)}\) (c) \(\mathrm{MgCO_{3}(s)} + 2\mathrm{H^+(aq)} \rightarrow \mathrm{Mg^{2+}(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\) (d) \(2\mathrm{NaAl(CO_{3})(OH)_{2}(s)} + 6\mathrm{H^+(aq)} \rightarrow 2\mathrm{Na^+(aq)} + 2\mathrm{Al^{3+}(aq)} + 4\mathrm{H_{2}O(l)} + 2\mathrm{CO_{2}(g)}\) (e) \(\mathrm{CaCO_{3}(s)} + 2\mathrm{H^+(aq)} \rightarrow \mathrm{Ca^{2+}(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\)

Step by step solution

01

1. Write the balanced chemical equation

Here are the balanced chemical equations for each reaction: (a) \(\mathrm{Al(OH)_{3}(s)} + 3\mathrm{HCl(aq)} \rightarrow \mathrm{AlCl_{3}(aq)} + 3\mathrm{H_{2}O(l)}\) (b) \(\mathrm{Mg(OH)_{2}(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl_{2}(aq)} + 2\mathrm{H_{2}O(l)}\) (c) \(\mathrm{MgCO_{3}(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl_{2}(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\) (d) \(2\mathrm{NaAl(CO_{3})(OH)_{2}(s)} + 6\mathrm{HCl(aq)} \rightarrow 2\mathrm{NaCl(aq)} + 2\mathrm{AlCl_{3}(aq)} + 4\mathrm{H_{2}O(l)} + 2\mathrm{CO_{2}(g)}\) (e) \(\mathrm{CaCO_{3}(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{CaCl_{2}(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\)
02

2. Separate aqueous ions

Now, we break down each aqueous solution into their respective ions: (a) \(\mathrm{Al(OH)_{3}(s)} + 3\mathrm{H^+(aq)} + 3\mathrm{Cl^-(aq)} \rightarrow \mathrm{Al^{3+}(aq)} + 3\mathrm{Cl^-(aq)} + 3\mathrm{H_{2}O(l)}\) (b) \(\mathrm{Mg(OH)_{2}(s)} + 2\mathrm{H^+(aq)} + 2\mathrm{Cl^-(aq)} \rightarrow \mathrm{Mg^{2+}(aq)} + 2\mathrm{Cl^-(aq)} + 2\mathrm{H_{2}O(l)}\) (c) \(\mathrm{MgCO_{3}(s)} + 2\mathrm{H^+(aq)} + 2\mathrm{Cl^-(aq)} \rightarrow \mathrm{Mg^{2+}(aq)} + 2\mathrm{Cl^-(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\) (d) \(2\mathrm{NaAl(CO_{3})(OH)_{2}(s)} + 6\mathrm{H^+(aq)} + 6\mathrm{Cl^-(aq)} \rightarrow 2\mathrm{Na^+(aq)} + 2\mathrm{Cl^-(aq)} + 2\mathrm{Al^{3+}(aq)} + 6\mathrm{Cl^-(aq)} + 4\mathrm{H_{2}O(l)} + 2\mathrm{CO_{2}(g)}\) (e) \(\mathrm{CaCO_{3}(s)} + 2\mathrm{H^+(aq)} + 2\mathrm{Cl^-(aq)} \rightarrow \mathrm{Ca^{2+}(aq)} + 2\mathrm{Cl^-(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\)
03

3. Cancel out spectator ions

We proceed to cancel out the spectator ions in each equation: (a) \(\cancel{3\mathrm{Cl^-(aq)}}\) (b) \(\cancel{2\mathrm{Cl^-(aq)}}\) (c) \(\cancel{2\mathrm{Cl^-(aq)}}\) (d) \(\cancel{2\mathrm{Cl^-(aq)}}\) and \(\cancel{6\mathrm{Cl^-(aq)}}\) (e) \(\cancel{2\mathrm{Cl^-(aq)}}\)
04

4. Write net ionic equations

Finally, we write the net ionic equations for each reaction: (a) \(\mathrm{Al(OH)_{3}(s)} + 3\mathrm{H^+(aq)} \rightarrow \mathrm{Al^{3+}(aq)} + 3\mathrm{H_{2}O(l)}\) (b) \(\mathrm{Mg(OH)_{2}(s)} + 2\mathrm{H^+(aq)} \rightarrow \mathrm{Mg^{2+}(aq)} + 2\mathrm{H_{2}O(l)}\) (c) \(\mathrm{MgCO_{3}(s)} + 2\mathrm{H^+(aq)} \rightarrow \mathrm{Mg^{2+}(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\) (d) \(2\mathrm{NaAl(CO_{3})(OH)_{2}(s)} + 6\mathrm{H^+(aq)} \rightarrow 2\mathrm{Na^+(aq)} + 2\mathrm{Al^{3+}(aq)} + 4\mathrm{H_{2}O(l)} + 2\mathrm{CO_{2}(g)}\) (e) \(\mathrm{CaCO_{3}(s)} + 2\mathrm{H^+(aq)} \rightarrow \mathrm{Ca^{2+}(aq)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}\)

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 91影视!

Key Concepts

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

Antacids
Antacids are essential medications used to neutralize excess stomach acid, providing relief from discomfort and healing minor ulcers. These substances work by neutralizing gastric acidity, primarily by increasing the pH level in the stomach. This reduction in acidity can alleviate symptoms such as heartburn, indigestion, and stomach pain. Some common compounds used in antacids include:
  • Aluminum hydroxide,
  • Magnesium hydroxide,
  • Calcium carbonate.
These compounds react with hydrochloric acid (HCl), which is the main component of stomach acid, to form salts and water. This neutralization reaction is fundamental in the functioning of antacids. Understanding the chemical process behind these reactions will help you appreciate how antacids alleviate symptoms associated with excess stomach acid.
Acid-Base Reactions
Acid-base reactions are chemical processes that involve the transfer of hydrogen ions (H^+). An acid is a substance that can donate a hydrogen ion, while a base is one that can accept it. When acids and bases react, they often produce a salt and water. In a typical antacid reaction, the base (a component of the antacid) reacts with the hydrochloric acid (HCl) in the stomach.For example, when magnesium hydroxide reacts with HCl, the reaction is as follows:\( \text{Mg(OH)}_2 + \text{2HCl} \rightarrow \text{MgCl}_2 + \text{2H}_2\text{O} \)The magnesium hydroxide acts as a base, accepting hydrogen ions from the hydrochloric acid, resulting in the formation of water, a neutral substance, and magnesium chloride, the salt. This type of reaction characterizes the neutralization process provided by antacids.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They show the substances involved in the reaction (reactants) and the substances produced (products). Balancing these equations is crucial because it respects the law of conservation of mass, meaning the number of atoms of each element must be the same on both sides of the equation.Consider the reaction between aluminum hydroxide and hydrochloric acid:\( \text{Al(OH)}_3 + \text{3HCl} \rightarrow \text{AlCl}_3 + \text{3H}_2\text{O} \)Here, the equation is balanced as each element has the same number of atoms on both the reactant and product sides. Balancing the chemical equation ensures that we accurately capture the process occurring on a molecular level during the reaction, allowing us to predict the outcome of the reaction precisely.
Spectator Ions
In chemical reactions that occur in aqueous solutions, not all ions participate directly in the reaction. Ions that do not change their oxidation state and remain in the solution unchanged are called spectator ions. They are present to maintain electrical neutrality but do not participate in the chemical change.When writing net ionic equations, spectator ions are removed since they do not influence the reaction. For example, in the reaction between magnesium hydroxide and hydrochloric acid:\( \text{Mg(OH)}_2 + \text{2H}^+ + 2\text{Cl}^- \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- + \text{2H}_2\text{O} \)Chloride ions (Cl^-) are spectator ions because they appear on both sides of the equation and do not affect the reaction significantly. By omitting them, we simplify the equation to reflect only the reacting ions, providing clearer insight into the essential changes.

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

We have seen that ions in aqueous solution are stabilized by the attractions between the ions and the water molecules. Why then do some pairs of ions in solution form precipitates? \([\) Section 4.2\(]\)

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\), has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with \(\mathrm{NaOH}\). It requires \(24.65 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) solution to titrate both acidic protons in \(50.00 \mathrm{~mL}\) of the tartaric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the tartaric acid solution.

A \(3.455-\mathrm{g}\) sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If \(0.2815 \mathrm{~g}\) of barium sulfate was obtained, what was the mass percentage of barium in the sample?

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation
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.