/*! 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 13 What is the pH of a 0.0015 M sol... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 13

What is the pH of a 0.0015 M solution of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\)

Short Answer

Expert verified
The pH of a 0.0015 M \(\mathrm{Ba(OH)}_2\) solution is 11.48.

Step by step solution

01

Determine the Dissociation of the Base

Barium hydroxide, \(\mathrm{Ba(OH)}_2\), is a strong base that dissociates completely in water. The dissociation can be represented by the equation:\[ \mathrm{Ba(OH)}_2 (aq) \rightarrow \mathrm{Ba}^{2+} (aq) + 2\mathrm{OH}^- (aq) \]Since each molecule of \(\mathrm{Ba(OH)}_2\) releases two hydroxide ions \((\mathrm{OH}^-)\), we need to account for that in our calculations.
02

Calculate Hydroxide Ion Concentration

Since each molecule of \(\mathrm{Ba(OH)}_2\) dissociates to produce two \(\mathrm{OH}^-\) ions, the concentration of \(\mathrm{OH}^-\) ions in the solution is twice the concentration of \(\mathrm{Ba(OH)}_2\). Therefore, the hydroxide ion concentration is:\[ \text{[OH}^-\text{]} = 2 \times 0.0015 \, \text{M} = 0.0030 \, \text{M} \]
03

Calculate pOH

The \(\text{pOH}\) of a solution is calculated using the formula: \[ \text{pOH} = -\log_{10}[\text{OH}^-] \]Substitute \([\text{OH}^-]\) with the value found in Step 2:\[ \text{pOH} = -\log_{10}(0.0030) \approx 2.52 \]
04

Calculate pH from pOH

The \(\text{pH}\) and \(\text{pOH}\) are related by the equation: \[ \text{pH} + \text{pOH} = 14 \]Using the \(\text{pOH}\) value calculated in Step 3, find \(\text{pH}\):\[ \text{pH} = 14 - 2.52 = 11.48 \]
05

Conclusion

The pH of a 0.0015 M solution of \(\mathrm{Ba(OH)}_2\) is 11.48.

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.

Strong Base
Barium hydroxide, or \(\mathrm{Ba(OH)}_2\), is classified as a strong base. This means it dissociates completely in water. When \(\mathrm{Ba(OH)}_2\) dissolves, every molecule separates into its ions completely:
  • Barium ions \((\mathrm{Ba}^{2+})\)
  • Hydroxide ions \((\mathrm{OH}^-)\)

This full dissociation is a key trait of strong bases as it ensures a higher availability of hydroxide ions in the solution, directly influencing the solution's pH.
Hydroxide Ion Concentration
In solutions of strong bases, the concentration of hydroxide ions \((\mathrm{OH}^-)\) can be calculated directly from the base's formula. For \(\mathrm{Ba(OH)}_2\):
Each molecule yields two hydroxide ions. If the solution concentration is \(0.0015\, \text{M}\), then:
  • The hydroxide ion concentration \([\text{OH}^-]\) is \(2 \times 0.0015\, \text{M} = 0.0030\, \text{M}\)

This doubling is specific to compounds that release more than one hydroxide ion per molecule.
pOH
The pOH is a measure of the concentration of hydroxide ions in a solution. It is calculated with the formula: \[ \text{pOH} = -\log_{10}[\text{OH}^-] \]
For our \(\mathrm{Ba(OH)}_2\) solution:
  • Substituting the hydroxide concentration: \(\text{pOH} = -\log_{10}(0.0030) \approx 2.52\)

This low pOH value indicates a basic solution. The more hydroxide present, the lower the pOH.
Water Dissociation
In water, pH and pOH are interconnected because of water's natural equilibrium. The equation:
  • \( \text{pH} + \text{pOH} = 14 \)

shows this relationship. Water naturally dissociates to form \(\text{H}^+\) and \(\text{OH}^-\), and this constant relationship holds for any aqueous solution.
By knowing the pOH, you can easily find the pH. For example, if \(\text{pOH} = 2.52\), then:
  • \( \text{pH} = 14 - 2.52 = 11.48 \)

This basic pH confirms the alkaline nature of a \(\mathrm{Ba(OH)}_2\) solution.

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

If \(K_{\mathrm{a}}\) for a weak acid is \(2.4 \times 10^{-11},\) what is the value of \(\mathrm{p} K_{\mathrm{a}} ?\)

Which of the following common food additives would give a basic solution when dissolved in water? (a) \(\mathrm{NaNO}_{3}\) (used as a meat preservative) (b) \(\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\) (sodium benzoate; used as a soft-drink preservative) (c) \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) (used as an emulsifier in the manufacture of pasteurized cheese)

A monoprotic acid HX has \(K_{a}=1.3 \times 10^{-3} .\) Calculate the equilibrium concentration of \(\mathrm{HX}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{pH}\) for a \(0.010 \mathrm{M}\) solution of the acid.

Decide whether each of the following substances should be classified as a Lewis acid or a Lewis base. (a) \(\mathrm{BCl}_{3}\) (Hint: Draw the electron dot structure.) (b) \(\mathrm{H}_{2} \mathrm{NNH}_{2},\) hydrazine (Hint: Draw the electron dot structure. (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (Hint: Draw the electron dot structure.)

You purchase a bottle of water. On checking its pH, you find that it is not neutral as you might have expected. Instead, it is slightly acidic. Why?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

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.