/*! 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 62 Determine the pH of each of the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the pH of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix D): (a) \(0.095 M\) hypochlorous acid, (b) \(0.0085 \mathrm{M}\) phenol, (c) \(0.095 \mathrm{M}\) hydroxylamine.

Short Answer

Expert verified
(a) For the 0.095 M Hypochlorous acid solution, determine the concentration of H+ ions using the given Ka value and the equation: \[Ka = \frac{x^2}{0.095 - x}\]. Then, calculate the pH using the formula: \[pH = -\log[H+]\]. (b) For the 0.0085 M Phenol solution, determine the concentration of H+ ions using the given Ka value and the equation: \[Ka = \frac{x^2}{0.0085 - x}\]. Then, calculate the pH using the formula: \[pH = -\log[H+]\]. (c) For the 0.095 M Hydroxylamine solution, determine the concentration of OH- ions using the given Kb value and the equation: \[Kb = \frac{x^2}{0.095 - x}\]. Calculate the pOH using the formula: \[pOH = -\log[OH^-]\]. Finally determine the pH using the relationship: \[pH = 14 - pOH\].

Step by step solution

01

(a) Hypochlorous Acid Solution

Hypochlorous acid (HClO) is a weak acid, so we will use its Ka value to find the concentration of H+ ions in the solution. Given, the concentration of Hypochlorous acid (HClO) = 0.095 M The dissociation reaction for HClO is: \[HClO \rightleftharpoons H+ + ClO^-\] For the given concentration, let x be the concentration of H+ ions at equilibrium, then the concentration of ClO- ions will also be x. So, \[Ka = \frac{[H+][ClO^-]}{[HClO]} = \frac{x^2}{0.095 - x}\] We are given the Ka value of HClO in Appendix D. Solve the equation for x to find the concentration of H+ ions, then use this value to calculate the pH using the formula: \[pH = -\log[H+]\]
02

(b) Phenol Solution

Phenol (C6H5OH) is a weak acid, so we will use its Ka value to find the concentration of H+ ions in the solution. Given, the concentration of Phenol (C6H5OH) = 0.0085 M The dissociation reaction for C6H5OH is: \[C6H5OH \rightleftharpoons H+ + C6H5O^-\] For the given concentration, let x be the concentration of H+ ions at equilibrium, then the concentration of C6H5O- ions will also be x. So, \[Ka = \frac{[H+][C6H5O^-]}{[C6H5OH]} = \frac{x^2}{0.0085 - x}\] We are given the Ka value of C6H5OH in Appendix D. Solve the equation for x to find the concentration of H+ ions, then use this value to calculate the pH using the formula: \[pH = -\log[H+]\]
03

(c) Hydroxylamine Solution

Hydroxylamine (NH2OH) is a weak base, so we will use its Kb value to find the concentration of OH- ions in the solution. Given, the concentration of hydroxylamine (NH2OH) = 0.095 M The dissociation reaction for NH2OH is: \[NH2OH + H2O \rightleftharpoons NH3OH+ + OH^-\] For the given concentration, let x be the concentration of OH- ions at equilibrium, then the concentration of NH3OH+ ions will also be x. So, \[Kb = \frac{[NH3OH+][OH^-]}{[NH2OH]} = \frac{x^2}{0.095 - x}\] We are given the Kb value of NH2OH in Appendix D. Solve the equation for x to find the concentration of OH- ions, and then calculate the pOH using the formula: \[pOH = -\log[OH^-]\] Finally, use the relationship between pH and pOH: \[pH = 14 - pOH\]

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weak Acids and Bases
Weak acids and bases differ from their strong counterparts in how they dissociate in water. While strong acids and bases completely dissociate into ions, weak acids and bases only partially do so. This partial dissociation means they establish equilibrium between their undissociated form and the ions they produce.
Some examples include:
  • Hypochlorous acid (\(\text{HClO}\))
  • Phenol (\(\text{C}_6\text{H}_5\text{OH}\))
  • Hydroxylamine (\(\text{NH}_2\text{OH}\))
In any solution, this equilibrium results in fewer ions being present compared to a strong acid or base solution of the same concentration. This is why weak acids and bases have higher pH values, indicating less acidity or basicity respectively.
Understanding the behavior of weak acids and bases is crucial in predicting the pH of their solutions. It involves recognizing that their incomplete dissociation leads to equilibrium states that we can describe mathematically.
Equilibrium Constants
The strength of a weak acid or base is quantified by its equilibrium constant. For acids, this is called the acid dissociation constant, \(K_a\), and for bases, it is the base dissociation constant, \(K_b\). These constants provide a measure of the tendency of a compound to dissociate into its constituent ions.
  • A larger \(K_a\) value indicates a stronger acid, as it implies a greater degree of dissociation into hydrogen ions (\([H^+]\)).
  • Similarly, a larger \(K_b\) value indicates a stronger base, as it shows a greater tendency to produce hydroxide ions (\([OH^-]\)).
To use these constants in calculations:
  • Set up an expression using the dissociation equation.
  • The expression relates the concentrations of products to reactants.
  • The values for \(K_a\) and \(K_b\) are typically found in tables or appendices.
These constants are essential for determining the extent of dissociation and thus determining the pH or pOH of a solution. Examining \(K_a\) and \(K_b\) helps predict the behavior of weak acids and bases.
Dissociation Reactions
Dissociation reactions describe the process by which a compound breaks down into its ions in solution. For weak acids like hypochlorous acid and phenol, the general dissociation reaction can be represented as:\[HA \rightleftharpoons H^+ + A^-\]Here, \(HA\) represents the undissociated acid, and \([H^+]\) and \([A^-]\) are the ions formed in solution.
For weak bases like hydroxylamine, the dissociation can be shown as:\[B + H_2O \rightleftharpoons BH^+ + OH^-\]In this reaction, \(B\) is the weak base, and it reacts with water to form hydroxide ions.
The concentration of these ions determines the solution's pH or pOH. The degree of dissociation is influenced by the equilibrium constant and the solution concentration, guiding the stoichiometry of the reaction in equilibrium.
Acid-Base Equilibria
Acid-base equilibria involve the balanced state between acids and bases and their ions in a solution. Understanding this concept is vital for predicting the pH value of solutions involving weak acids or bases.
To find the pH of weak acid solutions:
  • Identify the acid's dissociation constant \(K_a\).
  • Use the initial concentration to set up an equilibrium expression.\[K_a = \frac{[H^+][A^-]}{[HA]}\]
  • Solve for \([H^+]\), then calculate pH as \(-\log[H^+]\).
For weak base solutions:
  • Use the base's dissociation constant \(K_b\).
  • From the equilibrium expression:\[K_b = \frac{[BH^+][OH^-]}{[B]}\]
  • Find \([OH^-]\) and convert it to \(pOH\). Use \(pH = 14 - pOH\).
Acid-base equilibria ensure the understanding of how changes in concentrations or \(pK_a/pK_b\) values can shift balances, impacting the resulting pH of the solution. It’s a fundamental piece of chemical equilibrium theory, providing insights into the behavior of weak acids and bases in aqueous solutions.

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

Predict whether aqueous solutions of the following compounds are acidic, basic, or neutral: (a) \(\mathrm{NH}_{4} \mathrm{Br}\), (b) \(\mathrm{FeCl}_{3}\), (c) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), (d) \(\mathrm{KClO}_{4}\), (e) \(\mathrm{NaHC}_{2} \mathrm{O}_{4}\)

Predict whether aqueous solutions of the following substances are acidic, basic, or neutral: (a) \(\mathrm{CrBr}_{3}\), (b) LiI, (c) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) (d) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}\right] \mathrm{Cl}\), (e) \(\mathrm{KHSO}_{4}\)

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), received is name because it is present in sour milk as a product of bacterial action. It is also responsible for the soreness in muscles after vigorous exercise. (a) The \(\mathrm{pK}_{a}\) of lactic acid is \(3.85\). Compare this with the value for propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}, \mathrm{p} K_{a}=4.89\right)\), and explain the differ- ence. (b) Calculate the lactate ion concentration in a \(0.050 \mathrm{M}\) solution of lactic acid. (c) When a solution of sodium lactate, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COONa}\), is mixed with an aqueous copper(II) solution, it is possible to obtain a solid salt of copper(II) lactate as a blue-green hydrate, \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COO}\right)_{2} \mathrm{Cu} \cdot x \mathrm{H}_{2} \mathrm{O} .\) Elemental analysis of the solid tells us that the solid is \(22.9 \% \mathrm{Cu}\) and \(26.0 \% \mathrm{C}\) by mass. What is the value for \(x\) in the formula for the hydrate? (d) The acid-dissociation constant for the \(\mathrm{Cu}^{2+}(a q)\) ion is \(1.0 \times 10^{-8} .\) Based on this value and the acid-dissociation constant of lactic acid, predict whether a solution of copper(II) lactate will be acidic, basic, or neutral. Explain your answer.

What is the \(\mathrm{pH}\) of a solution that is \(2.5 \times 10^{-9} \mathrm{M}\) in \(\mathrm{NaOH}\) ? Does your answer make sense?

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.

See all solutions

Recommended explanations on Chemistry Textbooks

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.