Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 89

The active ingredients in a particular antacid tablet are aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3},\) and magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2} . \quad \mathrm{A} 5.00 \times 10^{2} \mathrm{mg}\) sample of the active ingredients was dissolved in \(50.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HCl} .\) The resulting solution, which was still acidic, required \(16.5 \mathrm{mL}\) of \(0.377 \mathrm{M} \mathrm{NaOH}\) for neutralization. What are the mass percentages of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) in the sample?

Short Answer

Expert verified
To solve this problem, the number of moles of H+ ions that were neutralized by the tablet needs to be calculated, followed by calculation of the number of moles of the active ingredients, and finally the mass percentages of the active ingredients. As it is not possible to solve \(3x + 2y = 0.025 - 0.006205 (x\geq0, y\geq0)\) without more information, the exact numerical answer cannot be given, but the procedure to find the masses and mass percentages is described.

Step by step solution

01

Calculate Hydrogen Ion Concentration

First, the total number of moles of H+ ions neutralized by both the active ingredients and NaOH needs to be calculated. This is done using the molarity and volume of HCl and NaOH. Molarity (\(M\)) is moles/liter, so moles = Molarity * Volume (in liters). For HCl, moles = \(0.500 M * 0.050 L = 0.025 mol\). For NaOH, moles = \(0.377 M * 0.0165 L = 0.006205 mol\).
02

Calculate Number of Moles of \(Al(OH)_3\) and \(Mg(OH)_2\)

Next, using stoichiometry, calculate the moles of the active ingredients that reacted with the H+ ions. This requires understanding that each mole of \(Al(OH)_{3}\) reacts with 3 moles of H+ ions and each mole of \(Mg(OH)_{2}\) reacts with 2 moles of H+ ions. Let the moles of \(Al(OH)_{3}\) be \(x\) and \(Mg(OH)_{2}\) be \(y\), then from the reaction it can be written as \(3x + 2y = 0.025 - 0.006205 (x\geq0, y\geq0)\). The left side of the equation represents the total amount of H+ ions, which must be equal to the total number of H+ ions neutralized (that is, the moles from HCl minus the moles from NaOH). While \(x\) and \(y\) are real numbers because number of moles cannot be negative.
03

Calculate the Mass Percentages

Convert the moles of \(Al(OH)_{3}\) and \(Mg(OH)_{2}\) to mass using their molar masses (Al = 26.98 g/mol, O = 16.00 g/mol, H = 1.01 g/mol, Mg = 24.31 g/mol). The molar masses of \(Al(OH)_{3}\) and \(Mg(OH)_{2}\) are 78.00 g/mol and 58.32 g/mol respectively. The masses are \(x * \) 78.00 g/mol for \(Al(OH)_{3}\) and \(y * \) 58.32 g/mol for \(Mg(OH)_{2}\). Then, calculate the mass percentage by dividing the mass of each compound by the total mass of the tablet and multiplying by 100. The mass percentage of \(Al(OH)_{3}\) and \(Mg(OH)_{2}\) are \(\frac{x * 78.00}{500} * 100\%\) and \(\frac{y * 58.32}{500} * 100\%\) respectively.

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.

Start your free trial

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

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

Sodium hydroxide used to make standard \(\mathrm{NaOH}(\mathrm{aq})\) solutions for acid-base titrations is invariably contaminated with some sodium carbonate. (a) Explain why, except in the most precise work, the presence of this sodium carbonate generally does not seriously affect the results obtained, for example, when \(\mathrm{NaOH}(\mathrm{aq})\) is used to titrate HCl(aq). (b) Conversely, show that if \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) comprises more than \(1 \%\) to \(2 \%\) of the solute in NaOH(aq), the titration results are affected.

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)

Blood alcohol content (BAC) is often reported in weight-volume percent (w/v\%). For example, a BAC of \(0.10 \%\) corresponds to \(0.10 \mathrm{g} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) per 100 mL of blood. Estimates of BAC can be obtained from breath samples by using a number of commercially available instruments, including the Breathalyzer for which a patent was issued to R. F. Borkenstein in 1958\. The chemistry behind the Breathalyzer is described by the oxidation- reduction reaction below, which occurs in acidic solution: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{g})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq}) \quad(\text { not balanced })\)A Breathalyzer instrument contains two ampules, each of which contains \(0.75 \mathrm{mg} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) dissolved in \(3 \mathrm{mL}\) of \(9 \mathrm{mol} / \mathrm{L} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) .\) One of the ampules is used as reference. When a person exhales into the tube of the Breathalyzer, the breath is directed into one of the ampules, and ethyl alcohol in the breath converts \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) into \(\mathrm{Cr}^{3+} .\) The instrument compares the colors of the solutions in the two ampules to determine the breath alcohol content (BrAC), and then converts this into an estimate of BAC. The conversion of BrAC into BAC rests on the assumption that 2100 mL of air exhaled from the lungs contains the same amount of alcohol as \(1 \mathrm{mL}\) of blood. With the theory and assumptions described in this problem, calculate the molarity of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in the ampules before and after a breath test in which a person with a BAC of \(0.05 \%\) exhales 0.500 Lof his breath into a Breathalyzer instrument.

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

Write a balanced equation for these redox reactions. (a) The oxidation of nitrite ion to nitrate ion by permanganate ion, \(\mathrm{MnO}_{4}^{-}\), in acidic solution \(\left(\mathrm{MnO}_{4}^{-}\right.\) ion is reduced to \(\mathrm{Mn}^{2+}\) ). (b) The reaction of manganese(II) ion and permanganate ion in basic solution to form solid manganese dioxide. (c) The oxidation of ethanol by dichromate ion in acidic solution, producing chromium(III) ion, acetaldehyde \(\left(\mathrm{CH}_{3} \mathrm{CHO}\right),\) and water as products.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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.

Sign-up for free