Chapter 13: Problem 15
Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in solutions with the following \(\mathrm{pH}\). (a) \(4.0\) (b) \(8.52\) (c) \(0.00\) (d) \(12.60\)
Short Answer
Expert verified
Given the pH values, calculate the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in the following solutions:
a) pH = 4.0
[H+] = \(1.0 \times 10^{-4}\) M, [OH-] = \(1.0 \times 10^{-10}\) M.
b) pH = 8.52
[H+] = \(2.91 \times 10^{-9}\) M, [OH-] = \(3.31 \times 10^{-6}\) M.
c) pH = 0.00
[H+] = \(1.0\) M, [OH-] = \(1.0 \times 10^{-14}\) M.
d) pH = 12.60
[H+] = \(2.51 \times 10^{-13}\) M, [OH-] = \(3.98 \times 10^{-2}\) M.
Step by step solution
01
Calculate [H+] using pH
Use the pH equation to find the concentration of hydrogen ions (H+):
pH = -log[H+] => [H+] = \(10^{-\mathrm{pH}}\) = \(10^{-4.0}\) = \(1.0 \times 10^{-4}\) M.
02
Calculate pOH
Since pH + pOH = 14, we can find the pOH:
pOH = 14 - pH = 14 - 4.0 = 10.0
03
Calculate [OH-] using pOH
Use the pOH equation to find the concentration of hydroxide ions (OH-):
pOH = -log[OH-] => [OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-10.0}\) = \(1.0 \times 10^{-10}\) M.
#a) Concentrations: [H+] = \(1.0 \times 10^{-4}\) M, [OH-] = \(1.0 \times 10^{-10}\) M.
#b) 8.52#
04
Calculate [H+] using pH
Use the pH equation to find the concentration of hydrogen ions (H+):
[H+] = \(10^{-\mathrm{pH}}\) = \(10^{-8.52}\) = \(2.91 \times 10^{-9}\) M.
05
Calculate pOH
Since pH + pOH = 14, we can find the pOH:
pOH = 14 - pH = 14 - 8.52 = 5.48
06
Calculate [OH-] using pOH
Use the pOH equation to find the concentration of hydroxide ions (OH-):
[OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-5.48}\) = \(3.31 \times 10^{-6}\) M.
#b) Concentrations: [H+] = \(2.91 \times 10^{-9}\) M, [OH-] = \(3.31 \times 10^{-6}\) M.
#c) 0.00#
07
Calculate [H+] using pH
Use the pH equation to find the concentration of hydrogen ions (H+):
[H+] = \(10^{-\mathrm{pH}}\) = \(10^{-0.0}\) = \(1.0\) M.
08
Calculate pOH
Since pH + pOH = 14, we can find the pOH:
pOH = 14 - pH = 14 - 0.0 = 14.0
09
Calculate [OH-] using pOH
Use the pOH equation to find the concentration of hydroxide ions (OH-):
[OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-14.0}\) = \(1.0 \times 10^{-14}\) M.
#c) Concentrations: [H+] = \(1.0\) M, [OH-] = \(1.0 \times 10^{-14}\) M.
#d) 12.60#
10
Calculate [H+] using pH
Use the pH equation to find the concentration of hydrogen ions (H+):
[H+] = \(10^{-\mathrm{pH}}\) = \(10^{-12.60}\) = \(2.51 \times 10^{-13}\) M.
11
Calculate pOH
Since pH + pOH = 14, we can find the pOH:
pOH = 14 - pH = 14 - 12.60 = 1.40
12
Calculate [OH-] using pOH
Use the pOH equation to find the concentration of hydroxide ions (OH-):
[OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-1.40}\) = \(3.98 \times 10^{-2}\) M.
#d) Concentrations: [H+] = \(2.51 \times 10^{-13}\) M, [OH-] = \(3.98 \times 10^{-2}\) M.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hydrogen Ion Concentration
The concentration of hydrogen ions, denoted as \([H^+]\), is a crucial measurement in determining a solution's acidity. The more hydrogen ions present, the more acidic the solution. This concentration is calculated from the \( ext{pH}\) of a solution using the formula: \([H^+] = 10^{-\text{pH}}\).
For example, if a solution has a \( ext{pH}\) of 4.0, we can calculate the hydrogen ion concentration as follows: \([H^+] = 10^{-4.0} = 1.0 \times 10^{-4}\) M.
This inverse logarithmic scale means that even small changes in \( ext{pH}\) result in significant changes in \([H^+]\).
It’s essential to comprehend that the \( ext{pH}\) scale typically ranges from 0 to 14, with each unit decrease corresponding to a tenfold increase in hydrogen ion concentration.
It's a measure for understanding how acidic a solution is:
For example, if a solution has a \( ext{pH}\) of 4.0, we can calculate the hydrogen ion concentration as follows: \([H^+] = 10^{-4.0} = 1.0 \times 10^{-4}\) M.
This inverse logarithmic scale means that even small changes in \( ext{pH}\) result in significant changes in \([H^+]\).
It’s essential to comprehend that the \( ext{pH}\) scale typically ranges from 0 to 14, with each unit decrease corresponding to a tenfold increase in hydrogen ion concentration.
It's a measure for understanding how acidic a solution is:
- A lower \( ext{pH}\) indicates a higher concentration of hydrogen ions (more acidic).
- A higher \( ext{pH}\) suggests a lower concentration of hydrogen ions (less acidic).
Hydroxide Ion Concentration
The concentration of hydroxide ions, \([OH^-]\), is an important parameter for determining the basicity of a solution. It is typically derived from the \( ext{pOH}\) of a solution, which can be found using the relation: \( ext{pOH} = 14 - \text{pH}\).
Once \( ext{pOH}\) is determined, the hydroxide ion concentration is calculated as \([OH^-] = 10^{-\text{pOH}}\).
For instance, if a solution has a \( ext{pH}\) of 8.52, the \( ext{pOH}\) would be calculated as follows: \( ext{pOH} = 14 - 8.52 = 5.48\).
Consequently, the hydroxide ion concentration can be derived: \([OH^-] = 10^{-5.48} = 3.31 \times 10^{-6}\) M.
Hydroxide ions are more prevalent in basic solutions, and they increase as the solution's \( ext{pH}\) rises above 7, which is the neutral midpoint.
Once \( ext{pOH}\) is determined, the hydroxide ion concentration is calculated as \([OH^-] = 10^{-\text{pOH}}\).
For instance, if a solution has a \( ext{pH}\) of 8.52, the \( ext{pOH}\) would be calculated as follows: \( ext{pOH} = 14 - 8.52 = 5.48\).
Consequently, the hydroxide ion concentration can be derived: \([OH^-] = 10^{-5.48} = 3.31 \times 10^{-6}\) M.
Hydroxide ions are more prevalent in basic solutions, and they increase as the solution's \( ext{pH}\) rises above 7, which is the neutral midpoint.
- A higher \( ext{pH}\) suggests a greater concentration of \([OH^-]\), indicating a more basic solution.
- A lower \( ext{pH}\) results in fewer hydroxide ions present.
pOH
The concept of \( ext{pOH}\) is closely related to \( ext{pH}\) and serves as a counterpart for basicity, focusing on hydroxide ion concentration in solutions.
Understanding \( ext{pOH}\) is essential since it completes the spectrum of acidity and basicity by covering the basic end of the scale. \( ext{pOH}\) is calculated using the relationship: \( ext{pH} + ext{pOH} = 14\)
For example, with a \( ext{pH}\) of 0.00, we can deduce that \( ext{pOH} = 14 - 0.00 = 14.0\).
Similarly, when you have a \( ext{pH}\) of 12.60, the calculation becomes \( ext{pOH} = 14 - 12.60 = 1.40\).
These calculations demonstrate the balance between acidity and basicity in water-derived solutions, illustrating how as one increases, the other decreases. Both \( ext{pH}\) and \( ext{pOH}\) help chemists understand the chemical properties of solutions such as their reactivity, taste, and corrosion potential.
Understanding \( ext{pOH}\) is essential since it completes the spectrum of acidity and basicity by covering the basic end of the scale. \( ext{pOH}\) is calculated using the relationship: \( ext{pH} + ext{pOH} = 14\)
For example, with a \( ext{pH}\) of 0.00, we can deduce that \( ext{pOH} = 14 - 0.00 = 14.0\).
Similarly, when you have a \( ext{pH}\) of 12.60, the calculation becomes \( ext{pOH} = 14 - 12.60 = 1.40\).
These calculations demonstrate the balance between acidity and basicity in water-derived solutions, illustrating how as one increases, the other decreases. Both \( ext{pH}\) and \( ext{pOH}\) help chemists understand the chemical properties of solutions such as their reactivity, taste, and corrosion potential.