Chapter 15: Problem 58
Calculate the \(\mathrm{pH}\) of each of the following given the molar hydroxide ion concentration: (a) \(\left[\mathrm{OH}^{-}\right]=0.000031 \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=0.00000000066 \mathrm{M}\)
Short Answer
Expert verified
(a) pH ≈ 9.49; (b) pH ≈ 4.82.
Step by step solution
01
Identify the Relationship between pH and pOH
In order to find the \(\text{pH}\), we first need to determine the \(\text{pOH}\). The relationship between \(\text{pH}\) and \(\text{pOH}\) is given by the equation: \(\text{pH} + \text{pOH} = 14\).
02
Calculate pOH from [OH-] for part (a)
Given \([\mathrm{OH}^{-}]=0.000031 \, \mathrm{M}\), compute \(\text{pOH}\): \(\text{pOH} = -\log[\mathrm{OH}^{-}] = -\log(0.000031)\). When calculated, \(\text{pOH} \approx 4.51\).
03
Calculate pH for part (a)
Using the relationship \(\text{pH} + \text{pOH} = 14\), find \(\text{pH}\): \(\text{pH} = 14 - \text{pOH} = 14 - 4.51\). Thus, \(\text{pH} \approx 9.49\).
04
Calculate pOH from [OH-] for part (b)
Given \([\mathrm{OH}^{-}]=0.00000000066 \mathrm{M}\), compute \(\text{pOH}\): \(\text{pOH} = -\log(0.00000000066)\). When calculated, \(\text{pOH} \approx 9.18\).
05
Calculate pH for part (b)
Using the relationship \(\text{pH} + \text{pOH} = 14\), find \(\text{pH}\): \(\text{pH} = 14 - \text{pOH} = 14 - 9.18\). Thus, \(\text{pH} \approx 4.82\).
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.
pOH
In chemistry, understanding how to calculate \( ext{pOH} \) is crucial when determining the acidity or basicity of a solution. The \( ext{pOH} \) is a measure of the hydroxide ion concentration in a solution and is related to \( ext{pH} \) through a simple relationship. This relationship is given by the equation: \( ext{pH} + ext{pOH} = 14 \). \( ext{pOH} \) is calculated using the formula \( ext{pOH} = - ext{log}[ ext{OH}^-] \). Here, \( [ ext{OH}^-] \) represents the molar concentration of hydroxide ions in the solution. Understanding \( ext{pOH} \) is important because it helps to determine the basicity of the solution. A low \( ext{pOH} \) value indicates a high concentration of hydroxide ions, suggesting a basic solution, while a higher \( ext{pOH} \) suggests a lower concentration of hydroxide ions, indicating a more acidic environment. Both \( ext{pH} \) and \( ext{pOH} \) provide insight into the nature of the solution and are used together to deliver a comprehensive view of its chemical properties.
hydroxide ion concentration
The concentration of hydroxide ions \( [ ext{OH}^-] \) in a solution is a key factor in determining the solution's basicity. A hydroxide ion ( ext{OH}^-) is formed when water molecules dissociate. This ion is a critical component in controlling the balance of acids and bases in a solution. Calculating \( [ ext{OH}^-] \) involves determining the molarity of the ions present in a given volume of solution. Molarity represents the number of moles of the solute in one liter of solution (mol/L). \( ext{OH}^- \) concentrations are typically given in molarity as well. A higher concentration of hydroxide ions usually indicates a more basic (alkaline) solution, while a lower concentration suggests a more neutral or acidic solution. Understanding \( [ ext{OH}^-] \) is essential because it plays a crucial role in the calculation of \( ext{pOH} \) and \( ext{pH} \). By knowing the concentration of hydroxide ions, we can use logarithmic calculations to find \( ext{pOH} \), and subsequently, determine the \( ext{pH} \) of the solution. This information is necessary in various scientific and industrial applications where the acidity or basicity level of a solution needs to be adjusted or monitored for optimal results.
logarithmic calculation
Logarithmic calculations are at the heart of \( ext{pH} \) and \( ext{pOH} \) computations. A logarithm, simply put, is an exponent or power to which a base, usually 10, must be raised to produce a given number. When we use \( ext{log} \) in chemistry, we typically refer to base 10 logarithms, also known as common logarithms. To find \( ext{pOH} \), we calculate the negative logarithm of the hydroxide ion concentration, expressed as \( ext{pOH} = - ext{log}[ ext{OH}^-] \). This form of logarithmic calculation simplifies working with very small or large numbers by expressing them in a compact form. The concept of using a negative logarithm ensures that positive values result when dealing with concentrations below 1. For instance, \( - ext{log}(0.000031) \) returns a positive \( ext{pOH} \) value, making it easy to handle and interpret mathematically and conceptually. Logarithms allow us to translate real-world exponential growth, or reduction in this instance, into a more comprehensible, linear numeric scale. For students and professionals regularly involved with chemical solutions, mastering logarithmic calculations offers a significant skill advantage in analyzing and understanding the chemical behavior of solutions.