/*! 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 89 What are the major species prese... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 89

What are the major species present in \(0.015 M\) solutions of each of the following bases? a. \(\mathrm{KOH}\) b. \(\mathrm{Ba}(\mathrm{OH})_{2}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Short Answer

Expert verified
For a $0.015 \, M$ KOH solution, the major species present are K鈦� and OH鈦�. The concentration of hydroxide ions [OH鈦籡 is 0.015 M. Calculate the pOH using the formula pOH = -log[OH鈦籡 and then find the pH using pH = 14 - pOH. For a $0.015 \, M$ Ba(OH)鈧� solution, the major species present are Ba虏鈦� and OH鈦�. As Ba(OH)鈧� produces 2 moles of OH鈦� ions per mole of the compound, [OH鈦籡 is twice the molarity of the Ba(OH)鈧� solution, which is $2 \times 0.015 \, M$. Calculate the pOH using the formula pOH = -log[OH鈦籡 and then find the pH using pH = 14 - pOH.

Step by step solution

01

Dissociate the strong base in water

KOH is a strong base and will dissociate completely in water, producing K鈦� and OH鈦� ions: \[ \mathrm{KOH} \overset{H_{2}O}{\longrightarrow} \mathrm{K^+} + \mathrm{OH^-} \] #a. KOH solution: [OH鈦籡 and pH calculation#
02

Find the [OH鈦籡 for KOH

The concentration of hydroxide ions [OH鈦籡 is equal to the molarity of the KOH solution. In this case, [OH鈦籡 = 0.015 M.
03

Calculate the pOH

To find the pOH, use the formula pOH = -log[OH鈦籡: \[ \mathrm{pOH} = -\log{(0.015)} \] Calculate the pOH value.
04

Calculate the pH

To find the pH, use the relationship pH = 14 - pOH. Plug in the pOH value from Step 3 to calculate the pH of the KOH solution. #b. Ba(OH)鈧� solution: Major species present#
05

Dissociate the strong base in water

Ba(OH)鈧� is a strong base and will dissociate completely in water, producing Ba虏鈦� and OH鈦� ions: \[ \mathrm{Ba(OH)_{2}} \overset{H_{2}O}{\longrightarrow} \mathrm{Ba^{2+}} + 2 \mathrm{OH^{-}} \] Note that Ba(OH)鈧� produces 2 moles of OH鈦� ions per mole of the compound. #b. Ba(OH)鈧� solution: [OH鈦籡 and pH calculation#
06

Find the [OH鈦籡 for Ba(OH)鈧�

Since Ba(OH)鈧� produces 2 moles of OH鈦� ions per mole of the compound, the concentration of hydroxide ions [OH鈦籡 is twice the molarity of the Ba(OH)鈧� solution: \[ [\mathrm{OH^-}] = 2 \times 0.015 \, M \] Calculate the [OH鈦籡 for the Ba(OH)鈧� solution.
07

Calculate the pOH

To find the pOH, use the formula pOH = -log[OH鈦籡: \[ \mathrm{pOH} = -\log{[2 \times 0.015]} \] Calculate the pOH value for the Ba(OH)鈧� solution.
08

Calculate the pH

To find the pH, use the relationship pH = 14 - pOH. Plug in the pOH value from Step 3 to calculate the pH of the Ba(OH)鈧� solution.

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.

Base Dissociation in Water
When strong bases like KOH and Ba(OH)2 are dissolved in water, they undergo a process known as base dissociation. This refers to the separation of the base into its constituent ions. For KOH, this dissociation is straightforward: one molecule of KOH yields one potassium ion (K+) and one hydroxide ion (OH-).

In contrast, barium hydroxide (Ba(OH)2) dissociates to produce one barium ion (Ba2+) and two hydroxide ions. The difference in the number of hydroxide ions produced is significant鈥攚hile KOH releases one OH- per molecule, Ba(OH)2 releases two, which impacts the pH calculation for these solutions.
Hydroxide Ion Concentration
The concentration of hydroxide ions, denoted as [OH-], is a critical aspect in calculating the pH of a basic solution. For a 0.015 M KOH solution, since KOH dissociates completely, the concentration of hydroxide ions is also 0.015 M. However, Ba(OH)2 is different due to its two hydroxide ions per formula unit, leading to a [OH-] that is twice the molarity of the Ba(OH)2 solution. Specifically, a 0.015 M Ba(OH)2 solution has an [OH-] of 0.030 M. Understanding these concentrations is necessary for accurately determining the solution's pH.
pOH and pH relationship
The concepts of pOH and pH are intrinsically linked through the equation pH + pOH = 14, which is valid at 25掳C. The pOH is found by taking the negative logarithm of the hydroxide ion concentration. Once pOH is determined, it can be subtracted from 14 to find the pH. This relationship highlights a fundamental principle of aqueous solutions: as the concentration of hydroxide ions increases, resulting in a lower pOH, the pH of the solution increases, indicating higher basicity.
Strong Base Properties
Strong bases, such as KOH and Ba(OH)2, exhibit distinct properties. They completely dissociate in water, leaving no undissociated molecules behind. This complete dissociation results in relatively high concentrations of hydroxide ions in solution as compared to weak bases which only partially dissociate. Therefore, the pH of strong base solutions will be relatively high, reflecting their strong basicity and their ability to raise the pH of the surrounding environment. Additionally, the ionic compounds of strong bases tend to be highly soluble in water, contributing to their ability to maintain high pH levels.

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

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a solution obtained by adding \(0.0100 \mathrm{~mol}\) solid \(\mathrm{NaOH}\) to \(1.00 \mathrm{~L}\) of \(15.0 \mathrm{M} \mathrm{NH}_{3}\).

Rank the following \(0.10 \mathrm{M}\) solutions in order of increasing \(\mathrm{pH}\). a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), \(\mathrm{HNO}_{3}\)

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables \(14.2\) and \(14.3 .\) a. \(\mathrm{KCl}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) e. \(\mathrm{NH}_{4} \mathrm{~F}\) b. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KF}\) f. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}\)

A \(10.0\) -mL sample of an \(\mathrm{HCl}\) solution has a \(\mathrm{pH}\) of \(2.000\). What volume of water must be added to change the \(\mathrm{pH}\) to \(4.000\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.