/*! 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 8 Calculate the \(\mathrm{pH}\) an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 8

Calculate the \(\mathrm{pH}\) and the \(\mathrm{pOH}\) of an aqueous solution that is \(0.020 \mathrm{M}\) in \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
pH = 12.60, pOH = 1.40

Step by step solution

01

Identify the Dissociation

Barium hydroxide, \( \text{Ba(OH)}_2 \), is a strong base that fully dissociates in water. The dissociation is given by the equation: \( \text{Ba(OH)}_2 (aq) \rightarrow \text{Ba}^{2+} (aq) + 2\text{OH}^- (aq) \). Each formula unit creates two hydroxide ions (\( \text{OH}^- \)).
02

Calculate Hydroxide Ion Concentration

Since \( \text{Ba(OH)}_2 \) dissociates completely, the concentration of \( \text{OH}^- \) ions is twice that of \( \text{Ba(OH)}_2 \), which is \( 0.020 \text{ M} \) initially. Therefore, the \( \text{OH}^- \) concentration is \( 2 \times 0.020 \text{ M} = 0.040 \text{ M} \).
03

Calculate pOH

The \( \text{pOH} \) of the solution can be calculated using the formula \( \text{pOH} = -\log{[\text{OH}^-]} \). Substitute the \( \text{OH}^- \) concentration: \( \text{pOH} = -\log{(0.040)} \approx 1.40 \).
04

Calculate pH from pOH

Using the relationship \( \text{pH} + \text{pOH} = 14 \) at \( 25^{\circ} \text{C} \), we calculate \( \text{pH} = 14 - 1.40 = 12.60 \).

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.

Ba(OH)2 dissociation
When barium hydroxide, represented as \( \text{Ba(OH)}_2 \), is added to water, it undergoes a process called dissociation. As a strong base, \( \text{Ba(OH)}_2 \) dissociates completely in water, leading to the formation of ions. The dissociation can be depicted by the equation: \[\text{Ba(OH)}_2 (aq) \rightarrow \text{Ba}^{2+} (aq) + 2\text{OH}^- (aq)\]This means that for each molecule of barium hydroxide, one barium ion \( \text{Ba}^{2+} \) and two hydroxide ions \( \text{OH}^- \) are produced.
  • Complete dissociation is a key characteristic of strong bases like \( \text{Ba(OH)}_2 \).
  • The production of two \( \text{OH}^- \) ions for every molecule doubles the hydroxide concentration in the solution.
Understanding this concept is crucial as it lays the foundation for calculating further chemical properties like pH.
hydroxide ion concentration
Given the complete dissociation of \( \text{Ba(OH)}_2 \), determining the hydroxide ion concentration involves straightforward multiplication. We start with a 0.020 M solution of \( \text{Ba(OH)}_2 \). Since each formula unit produces two \( \text{OH}^- \), this concentration effectively doubles. Therefore, the concentration of \( \text{OH}^- \) ions becomes twice the initial concentration of \( \text{Ba(OH)}_2 \).
  • Initial concentration of \( \text{Ba(OH)}_2 = 0.020 \text{ M} \)
  • \( \text{OH}^- \) concentration = \( 2 \times 0.020 \text{ M} = 0.040 \text{ M} \)
By understanding this multiplication, one can see how the presence of barium hydroxide impacts the ion balance and eventual pH levels of a solution.
pOH calculation
Once we have the hydroxide ion concentration, we can calculate the \( \text{pOH} \) of the solution. The formula used here is:\[\text{pOH} = -\log{[\text{OH}^-]}\]In our scenario, substituting the known \( \text{OH}^- \) concentration value gives us:\[\text{pOH} = -\log{(0.040)} \approx 1.40\]
  • The logarithm function here helps to convert ion concentration to a more manageable number.
  • As \( \text{OH}^- \) concentrations rise, \( \text{pOH} \) values decrease indicating stronger basicity.
Understanding \( \text{pOH} \) is crucial because it gives direct insight into the solution's basic properties, complementing the \( \text{pH} \) value.
acid-base equilibrium
In aqueous solutions, pH and pOH are interconnected, following the relationship:\[\text{pH} + \text{pOH} = 14\]This equation holds true at \( 25^{\circ} \text{C} \), where the scale of acidity and basicity is set. Given a \( \text{pOH} \) of 1.40, we can deduce:\[\text{pH} = 14 - 1.40 = 12.60\]
  • This relationship ensures that as one value increases, the other decreases.
  • A high \( \text{pH} \), as we calculated, indicates a strongly basic solution.
Comprehending this equilibrium helps in understanding how pH and pOH values provide insights into the solution鈥檚 overall acidic or basic nature, enabling us to predict the chemical behavior of substances in 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

Nitrites, such as \(\mathrm{NaNO}_{2}\), are added to processed meats and hamburger both as a preservative and to give the meat a redder color by binding to hemoglobin in the red blood cells. When the nitrite ion is ingested, it reacts with stomach acid to form nitrous acid, \(\mathrm{HNO}_{2}(a q)\), in the stomach. Given that \(K_{\mathrm{a}}=5.6 \times 10^{-4} \mathrm{M}\) for nitrous acid at \(25^{\circ} \mathrm{C}\), compute the value of the ratio \(\left[\mathrm{HNO}_{2}\right] /\left[\mathrm{NO}_{2}^{-}\right]\) in the stomach following ingestion of \(\mathrm{NaNO}_{2}(a q)\) when \(\left[\mathrm{H}_{3} \mathrm{O}^{4}\right]\) is \(0.10 \mathrm{M}\). Assume a temperature of \(25^{\circ} \mathrm{C}\).

The \(\mathrm{pH}\) of a \(0.050\) -M aqueous solution of pyruvic acid, \(\mathrm{CH}_{3} \mathrm{COCOOH}(a q)\), an intermediate in the metabolism of glucose, is found to be 1.91. Calculate \(K_{\mathrm{a}}\), the acid- dissociation constant, for pyruvic acid.

Some soaps are produced by reacting caustic soda, \(\mathrm{NaOH}(s)\), with animal fats, which contain a type of organic acid called a fatty acid. An example is stearic acid, whose chemical formula is \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COOH}\). Fatty acids, like most organic acids, are weak acids. Is a soap solution acidic, basic, or neutral?

Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(3.00\) -M acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}(a q)\), solution. Calculate the \(\mathrm{pH}\) of the resulting solution when \(2.00\) milliliters of the \(3.00-\mathrm{M}\) acetic acid is diluted to make a \(250.0\) -milliliter solution. The value of \(K_{\mathrm{a}}\) for acetic acid at \(25^{\circ} \mathrm{C}\) is \(1.8 \times 10^{-5} \mathrm{M}\)

Calculate the \(\mathrm{pH}\) and the \(\mathrm{pOH}\) of an aqueous solution that is \(0.035 \mathrm{M}\) in \(\mathrm{HCl}(a q)\) and \(0.045 \mathrm{M}\) in \(\mathrm{HBr}(a q)\) at \(25^{\circ} \mathrm{C}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical 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.