/*! 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 110 Calculate the \(\mathrm{pH}\) of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 110

Calculate the \(\mathrm{pH}\) of a solution made by adding \(2.50 \mathrm{~g}\) of lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right)\) to enough water to make \(1.500 \mathrm{~L}\) of solution.

Short Answer

Expert verified
The pH of the solution made by adding 2.50 g of lithium oxide to enough water to make 1.500 L of solution is approximately \(13.05\).

Step by step solution

01

Write the chemical equation for the reaction of Li2O with water

The chemical equation for the reaction of lithium oxide with water is given below: \[ \mathrm{Li}_{2}\mathrm{O(s)} + \mathrm{H}_{2}\mathrm{O(l)} \longrightarrow 2\mathrm{LiOH(aq)} \] In this reaction, each mole of Li2O reacts with water to produce 2 moles of lithium hydroxide (LiOH), which is a strong base and dissociates completely in water.
02

Calculate the number of moles of Li2O added to the water

First, let's find the molar mass of Li2O: Molar mass of Li2O = 2脳(Molar mass of Li) + Molar mass of O = 2脳(6.94 g/mol) + 16.00 g/mol = 29.88 g/mol Now, we can calculate the number of moles of Li2O: Number of moles (n) = mass / molar mass n(Li2O) = 2.50 g / 29.88 g/mol = 0.0837 mol
03

Determine the moles of OH- ions that will form from the moles of Li2O in the solution

Since one mole of Li2O produces two moles of LiOH, and each mole of LiOH produces one mole of OH- ions, the number of moles of OH- ions in the solution would be: n(OH-) = 2 脳 n(Li2O) = 2 脳 0.0837 mol = 0.1674 mol
04

Calculate the concentration of OH- ions

Now, we can calculate the concentration of OH- ions in the 1.500 L solution: Concentration (C) = moles/volume C(OH-) = 0.1674 mol / 1.500 L = 0.1116 M
05

Calculate the pOH of the solution

Now, we will calculate the pOH by taking the negative base-10 logarithm of the OH- concentration: pOH = -log10(C(OH-)) pOH = -log10(0.1116) = 0.951
06

Determine the pH of the solution by using the relation between pH and pOH

To calculate the pH of the solution, we need to use the relationship between pH and pOH: pH + pOH = 14 Now, we can find the pH: pH = 14 - pOH = 14 - 0.951 = 13.049 The pH of the solution is approximately 13.05.

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!

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

Which of the following solutions has the higher pH? (a) a \(0.1 \mathrm{M}\) solution of a strong acid or a \(0.1 \mathrm{M}\) solution of a weak acid, (b) a \(0.1 \mathrm{M}\) solution of an acid with \(K_{a}=2 \times 10^{-3}\) or one with \(K_{a}=8 \times 10^{-6}\), (c) a 0.1 M solution of a base with \(\mathrm{pK}_{b}=4.5\) or one with \(\mathrm{pK}_{b}=6.5\).

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) A \(0.035 \mathrm{M}\) solution of ephedrine has a pH of \(11.33 .\) (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}\), and \(\mathrm{OH}^{-} ?\) (b) Calculate \(K_{b}\) for ephedrine.

If a substance is a Lewis acid, is it necessarily a Br酶nsted-Lowry acid? Is it necessarily an Arrhenius acid? Explain.

Deuterium oxide \(\left(\mathrm{D}_{2} \mathrm{O}\right.\), where \(\mathrm{D}\) is deuterium, the hydrogen- 2 isotope) has an ion-product constant, \(K_{\mathrm{uu}}\) of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\) Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure(neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

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.