Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 117

A \(50.00-\mathrm{mL}\) sample of aqueous \(\mathrm{Ca}(\mathrm{OH})_{2}\) requires \(34.66 \mathrm{~mL}\) of a \(0.944-M\) nitric acid for neutralization. Calculate the concentration (molarity) of the original solution of calcium hydroxide.

Short Answer

Expert verified
The concentration of the original calcium hydroxide solution is \(0.327\mathrm{~M}\).

Step by step solution

01

Determine the moles of nitric acid used

The volume and concentration of nitric acid solution are given, so we can calculate the moles of nitric acid used to neutralize the calcium hydroxide solution. Use the formula: Moles \(= Volume \times Concentration\) Moles of nitric acid \(= 34.66 \mathrm{~mL} \times 0.944 \mathrm{\frac{mol}{L}}\ = 0.0327 \mathrm{mol}\)
02

Use stoichiometry to determine moles of calcium hydroxide present

In the balanced chemical equation, we observe that 2 moles of HNO3 (nitric acid) react with 1 mole of Ca(OH)2 (calcium hydroxide). Thus, we can find the moles of calcium hydroxide by dividing the moles of nitric acid by 2: Moles of calcium hydroxide \(= \frac{0.0327 \mathrm{~mol~of~HNO_3}}{2}\ = 0.01635\mathrm{~mol~of~Ca(OH)_2}\)
03

Calculate the concentration of the original calcium hydroxide solution

Now that we have the moles of calcium hydroxide present in the 50.00 mL sample, we can calculate the concentration of the original calcium hydroxide solution using the formula: Concentration \(= \frac{moles}{volume}\) Concentration of calcium hydroxide \(= \frac{0.01635\mathrm{~mol}}{50.00 \mathrm{~mL}}\ = \frac{0.01635\mathrm{~mol}}{0.05000 \mathrm{~L}}\ = 0.327\mathrm{\frac{mol}{L}}\) The concentration of the original calcium hydroxide solution is \(0.327\mathrm{~M}\).

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

A \(0.500-\mathrm{L}\) sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{~mL}\) of \(0.213 \mathrm{M} \mathrm{NaOH}\). After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{~mL}\) of \(0.103 M\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. \(0.100\) mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{~mL}\) of solution b. \(2.5\) moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(1.25 \mathrm{~L}\) of solution c. \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{~mL}\) of solution d. \(1.00 \mathrm{~g} \mathrm{~K}_{2} \mathrm{PO}_{4}\) in \(250.0 \mathrm{~mL}\) of solution

Which of the following solutions of strong electrolytes contains the largest number of ions: \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\), \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{BaCl}_{2}\), or \(75.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) ?

Polychlorinated biphenyls (PCBs) have been used extensively as dielectric materials in electrical transformers. Because PCBs have been shown to be potentially harmful, analysis for their presence in the environment has become very important. PCBs are manufactured according to the following generic reaction: $$ \mathrm{C}_{12} \mathrm{H}_{10}+{ }_{n \mathrm{Cl}_{2}} \rightarrow \mathrm{C}_{12} \mathrm{H}_{10-n} \mathrm{Cl}_{n}+n \mathrm{HCl} $$ This reaction results in a mixture of \(\mathrm{PCB}\) products. The mixture is analyzed by decomposing the \(\mathrm{PCBs}\) and then precipitating the resulting \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). a. Develop a general equation that relates the average value of \(n\) to the mass of a given mixture of \(\mathrm{PCBs}\) and the mass of \(\mathrm{AgCl}\) produced. b. A \(0.1947-g\) sample of a commercial PCB yielded \(0.4791 \mathrm{~g}\) of \(\mathrm{AgCl}\). What is the average value of \(n\) for this sample?

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) d. As \(_{4}\) e. \(\mathrm{HAsO}_{2}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

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