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

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 aqueous \(\mathrm{HCl}\) in the stomach and each of the following substances used in various antacids: (a) \(\mathrm{Al}(\mathrm{OH})_{3}(s),(\mathbf{b}) \mathrm{Mg}(\mathrm{OH})_{2}(s),(\mathbf{c}) \mathrm{MgCO}_{3}(s),\) (d) \(\mathrm{NaAl}\left(\mathrm{CO}_{3}\right)(\mathrm{OH})_{2}(s)\) (e) \(\mathrm{CaCO}_{3}(s)\)

Short Answer

Expert verified
(a) Al(OH)3(s) + 3H+(aq) → 3H2O(l) (b) Mg(OH)2(s) + 2H+(aq) → 2H2O(l) (c) CO3^2-(aq) + 2H+(aq) → H2O(l) + CO2(g) (d) CO3^2-(aq) + 2OH-(aq) + 6H+(aq) → 2H2O(l) + 3CO2(g) + Na+(aq) (e) CO3^2-(aq) + 2H+(aq) → H2O(l) + CO2(g)

Step by step solution

01

(a) Reaction between HCl and Al(OH)3

First, we will write the balanced molecular equation for the reaction between Al(OH)3 and HCl: Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l) To write the net ionic equation, let's break down the molecular equation into its ionic components: Al(OH)3(s) → Al^3+ (aq) + 3OH- (aq) 3HCl(aq) → 3H+ (aq) + 3Cl- (aq) AlCl3(aq) → Al^3+ (aq) + 3Cl- (aq) Since Al^3+(aq) and 3Cl-(aq) are the same on both sides of the equation, they are called spectator ions and can be eliminated from the net ionic equation. The net ionic equation for this reaction will be: Al(OH)3(s) + 3H+ (aq) → 3H2O(l)
02

(b) Reaction between HCl and Mg(OH)2

First, we will write the balanced molecular equation for the reaction between Mg(OH)2 and HCl: Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) Now, let's break down the molecular equation into its ionic components: Mg(OH)2(s) → Mg^2+ (aq) + 2OH- (aq) 2HCl(aq) → 2H+ (aq) + 2Cl- (aq) MgCl2(aq) → Mg^2+ (aq) + 2Cl- (aq) The spectator ions in this case are Mg^2+(aq) and 2Cl-(aq). The net ionic equation for this reaction will be: Mg(OH)2(s) + 2H+ (aq) → 2H2O(l)
03

(c) Reaction between HCl and MgCO3

First, we will write the balanced molecular equation for the reaction between MgCO3 and HCl: MgCO3(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g) Now, let's break down the molecular equation into its ionic components: MgCO3(s) → Mg^2+ (aq) + CO3^2- (aq) 2HCl(aq) → 2H+ (aq) + 2Cl- (aq) MgCl2(aq) → Mg^2+ (aq) + 2Cl- (aq) The spectator ions in this case are Mg^2+(aq) and 2Cl-(aq). The net ionic equation for this reaction will be: CO3^2- (aq) + 2H+ (aq) → H2O(l) + CO2(g)
04

(d) Reaction between HCl and NaAl(CO3)(OH)2

First, we will write the balanced molecular equation for the reaction between NaAl(CO3)(OH)2 and HCl: NaAl(CO3)(OH)2(s) + 6HCl(aq) → AlCl3(aq) + NaCl(aq) + 2H2O(l) + 3CO2(g) Now, let's break down the molecular equation into its ionic components: NaAl(CO3)(OH)2(s) → Na+ (aq) + Al^3+ (aq) + CO3^2- (aq) + 2OH- (aq) 6HCl(aq) → 6H+ (aq) + 6Cl- (aq) AlCl3(aq) → Al^3+ (aq) + 3Cl- (aq) NaCl(aq) → Na+ (aq) + Cl- (aq) The spectator ions in this case are Al^3+(aq) and Cl-(aq). The net ionic equation for this reaction will be: CO3^2- (aq) + 2OH- (aq) + 6H+ (aq) → 2H2O(l) + 3CO2(g) + Na+ (aq)
05

(e) Reaction between HCl and CaCO3

First, we will write the balanced molecular equation for the reaction between CaCO3 and HCl: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) Now, let's break down the molecular equation into its ionic components: CaCO3(s) → Ca^2+ (aq) + CO3^2- (aq) 2HCl(aq) → 2H+ (aq) + 2Cl- (aq) CaCl2(aq) → Ca^2+ (aq) + 2Cl- (aq) The spectator ions in this case are Ca^2+(aq) and 2Cl-(aq). The net ionic equation for this reaction will be: CO3^2- (aq) + 2H+ (aq) → H2O(l) + CO2(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 Vaia!

Key Concepts

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

Acid-Base Reactions
In acid-base reactions, acids react with bases to produce water and a salt. A good example of this is when hydrochloric acid (HCl), an acid present in the stomach, reacts with bases such as Mg(OH)_2 or Al(OH)_3. These reactions help neutralize excess stomach acid.
  • Acid: Substances that increase the concentration of H⁺ ions in a solution.
  • Base: Substances that increase the concentration of OH⁻ ions.
To identify the net ionic equation, it's crucial to focus on the ions that directly participate in forming water, leaving out spectator ions that do not take part in the reaction. When Al(OH)_3 reacts with HCl, the net ionic equation is:\[\mathrm{Al(OH)_3 (s) + 3H^+ (aq) \to 3H_2O (l)}\]Here, OH⁻ ions combine with H⁺ ions to form H_2O, demonstrating a fundamental principle of acid-base chemistry.
Antacids
Antacids are medications that help neutralize stomach acid, providing relief from symptoms like indigestion and heartburn. These agents often contain compounds like Al(OH)_3, Mg(OH)_2, or CaCO_3. They work by buffering the acidic environment of the stomach.
The role of antacids is to bring balance by reducing the acidity level. They undergo reactions that increase the pH of the stomach contents, making it less acidic. For instance:
  • \( \mathrm{Mg(OH)_2 (s) + 2H^+ (aq) \to 2H_2O (l)} \)
  • \( \mathrm{CaCO_3 (s) + 2H^+ (aq) \to H_2O (l) + CO_2 (g)} \)
In these reactions, the OH⁻ from the bases neutralizes the H⁺ ions in the stomach acid, forming water. The reaction with calcium carbonate (CaCO_3) also releases carbon dioxide gas, which can sometimes cause belching.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in a stable state where the concentrations of reactants and products remain constant.
In the context of acid-base reactions with antacids, reaching equilibrium means the antacid has effectively neutralized the stomach acid to where any additional acid doesn’t further react or change pH levels significantly. This concept helps to understand why only a specific amount of an antacid is needed to reach therapeutic effects. Dynamic Equilibrium in Reactions:
  • The rates of the forward and backward reactions become equal.
  • Concentrations of reactants and products stay constant but not necessarily equal.
For antacid reactions, once enough base has reacted to neutralize the acid, the system reaches an equilibrium where H⁺ ions are continuously formed and neutralized at the same rate, maintaining a balanced pH. Understanding equilibrium is key to correctly dosing antacids and ensuring effective relief without overuse.

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

You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out \(\mathrm{AgCl}(s) .\) What volume of a \(0.150 \mathrm{MHCl}(a q)\) solution is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of a \(0.200 \mathrm{MAgNO}_{3}\) solution? (b) You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (c) Given that a \(0.150 \mathrm{MHCl}(a q)\) solution costs \(\$ 39.95\) for \(500 \mathrm{~mL}\) and that KCl costs \(\$ 10 /\) ton, which analysis procedure is more cost-effective?

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 \(0.138 \mathrm{M}\), assuming no change in blood volume?

True or false: (a) If a substance is oxidized, there must be more oxygen in the substance. (b) If a substance is oxidized, it must lose at least one electron and form an anion.

Specify what ions are present upon dissolving each of the following substances in water: (a) \(\mathrm{HIO}_{3},\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2},\) (c) HCN, (d) \(\mathrm{CuSO}_{4}\).

Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions. (a) \(\mathrm{P}_{4}(s)+10 \mathrm{HClO}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Br}_{2}(l)+2 \mathrm{~K}(s) \longrightarrow 2 \mathrm{KBr}(s)\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)\) (d) \(\mathrm{ZnCl}_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+\)

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