/*! 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 93 A solution is \(0.18 M \mathrm{K... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 93

A solution is \(0.18 M \mathrm{KOH}\). What are the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\) in this solution?

Short Answer

Expert verified
[OH鈦籡 = 0.18 M, [H鈧僌鈦篯 鈮� 5.56 脳 10鈦宦光伌 M.

Step by step solution

01

Understand the Concept

The solution is described by its molarity of potassium hydroxide, KOH, which is a strong base. As a strong base, KOH dissociates completely in water into K鈦� and OH鈦� ions. We need to find the hydroxide ion concentration, [OH鈦籡, and from that calculate the hydronium ion concentration, [H鈧僌鈦篯, using the water ion product equilibrium.
02

Calculate [OH鈦籡 Concentration

Since KOH dissociates completely, its molarity equals the hydroxide ion concentration. Therefore, \[[\mathrm{OH}^{-}] = 0.18 \, \text{M}\]
03

Use Water Ion Product to Find [H鈧僌鈦篯

The ion product of water at 25掳C is a constant:\[K_w = [\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}] = 1.0 \times 10^{-14}\]We can rearrange this to find [H鈧僌鈦篯:\[[\mathrm{H}_{3}\mathrm{O}^{+}] = \frac{1.0 \times 10^{-14}}{[\mathrm{OH}^{-}]}\]Substitute [OH鈦籡 = 0.18 M into the equation:\[[\mathrm{H}_{3}\mathrm{O}^{+}] = \frac{1.0 \times 10^{-14}}{0.18 \, \text{M}} \approx 5.56 \times 10^{-14} \, \text{M}\]
04

Review and Confirm the Solution

Reassess the calculations to ensure they were performed correctly and align with the principles of chemistry. The [OH鈦籡 is correctly linked to the molarity of KOH, and the computation for [H鈧僌鈦篯 using the water ionization constant (Kw) is correct, maintaining significant figures appropriately for the initial given concentration.

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.

Molarity
Molarity is a fundamental concept in chemistry that represents the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solution, expressed as mol/L or simply M.
This unit simplifies calculations in chemistry, especially in reactions where precise concentrations are necessary for predicting chemical behavior.

  • To calculate molarity, divide the moles of solute by the liters of solution.
  • Molarity allows chemists to easily scale reactions and predict outcomes based on concentration.
For example, in the exercise, the molarity of the KOH solution is given as 0.18 M, meaning there are 0.18 moles of KOH in every liter of solution.
Strong Base
A strong base is a substance that completely dissociates into ions when dissolved in water. This means it breaks down fully to increase the concentration of hydroxide ions ( OH鈦� ) in the solution.
Potassium hydroxide (KOH) is a classic strong base. When KOH is dissolved in water, it dissociates completely into K鈦� and OH鈦� ions.

  • This complete dissociation is crucial because it determines the concentration of hydroxide ions in the solution.
  • In the given exercise, the 0.18 M KOH solution will have a hydroxide ion concentration of 0.18 M, due to complete dissociation.
Understanding the behavior of strong bases helps predict the changes they induce in the pH of water.
Water Ion Product
The water ion product, represented by the symbol K_w , is important in acid-base chemistry because it describes the equilibrium between hydrogen ions (H鈦�) and hydroxide ions (OH鈦�) in water.
At 25掳C, K_w is always constant and equals 1.0 imes 10^{-14} . This means that the product of the concentrations of H鈧僌鈦� and OH鈦� in pure water at this temperature always equates to 1.0 imes 10^{-14} .

  • The value of K_w allows us to calculate either H鈧僌鈦� or OH鈦� if one concentration is known.
  • This equilibrium is essential for calculating pH and solving problems like the one where H鈧僌鈦� concentration is found given OH鈦� .
In the example provided, using the known OH鈦� concentration helps solve for H鈧僌鈦� by rearranging the K_w expression.
Hydroxide Ion Concentration
The hydroxide ion concentration ( [OH鈦籡 ) is a key player in determining the basicity of a solution. In the case of strong bases like KOH, the hydroxide ion concentration is equal to the molarity of the solution because the base dissociates completely.
Knowing this concentration allows us to compute the hydronium ion concentration ( [H鈧僌鈦篯 ) using the water ion product.

  • The hydroxide ion concentration is used in the K_w equation to find hydronium ion concentration under equilibrium conditions.
  • For the provided exercise, with a hydroxide concentration of 0.18 M, the hydronium ion concentration is found using K_w .
By understanding the concentrations of ions, one can determine the acidity or basicity of a solution, and further predict the pH value.

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

Give the conjugate acid to each of the following species regarded as bases. a \(\mathrm{HS}^{-}\) b \(\mathrm{NH}_{2}\) c \(\mathrm{BrO}_{2}^{-}\) d) \(\mathrm{N}_{2} \mathrm{H}_{4}\)

Ethanol (ethyl alcohol), \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) can act as a Br酶nsted-Lowry acid. Write the chemical equation for the reaction of ethanol as an acid with hydroxide ion, \(\mathrm{OH}^{-}\). Ethanol can also react as a Br酶nsted-Lowry base. Write the chemical equation for the reaction of ethanol as a base with hydronium ion, \(\mathrm{H}_{3} \mathrm{O}^{+} .\) Explain how you arrived at these chemical equations. Both of these reactions can also be considered Lewis acid-base reactions. Explain this.

Give the conjugate base to each of the following species regarded as acids. a) \(\mathrm{HPO}_{4}^{2-}\) b) \(\mathrm{H}_{2} \mathrm{~S}\) c) HNO d) \(\mathrm{H}_{2} \mathrm{~A} \mathrm{~s} \mathrm{O}_{4}^{-}\)

For each of the following, write the complete chemical equation for the acid- base reaction that occurs. Describe each using Br酶nsted language (if appropriate) and then using Lewis language (show electron-dot formulas). a The \(\mathrm{ClO}^{-}\) ion reacts with water. b) The reaction of \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NH}_{2}^{-}\) in liquid ammonia to produce \(\mathrm{NH}_{3}\).

A solution of lye (sodium hydroxide, \(\mathrm{NaOH}\) ) has a hydroxide-ion concentration of \(0.050 M\). What is the pH at \(25^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

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.