Chapter 8: Problem 57
Find \(\mathrm{pH}\) and \(\mathrm{pOH}\) of each solution. a. \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-5} \mathrm{M}\) b. \(\left[\mathrm{H}^{+}\right]=2.8 \times 10^{-11} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-2} \mathrm{M}\) d. \(\left[\mathrm{OH}^{-}\right]=4.4 \times 10^{-9} \mathrm{M}\)
Short Answer
Step by step solution
Calculate pH from H+ concentration for (a) and (b)
Calculate pOH from pH using relation for (a) and (b)
Calculate pOH from OH- concentration for (c) and (d)
Calculate pH from pOH using relation for (c) and (d)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hydrogen Ion Concentration
To determine the \( ext{pH}\) from the hydrogen ion concentration, use the formula:
\[ ext{pH} = -\log[\text{H}^+]\]
This formula tells us that the \( ext{pH}\) is the negative logarithm (base 10) of the hydrogen ion concentration.
Points to remember:
- Lower \( ext{pH}\) values indicate higher acidity, meaning more hydrogen ions.
- A \( ext{pH}\) of 7 is neutral, typical for pure water.
- Higher \( ext{pH}\) values indicate basicity, meaning fewer hydrogen ions.
Hydroxide Ion Concentration
To find the \( ext{pOH}\) from the hydroxide ion concentration, use the formula:
\[ ext{pOH} = -\log[\text{OH}^-]\]
This formula calculates the pOH, which is the negative logarithm (base 10) of the hydroxide ion concentration.
Key points to remember:
- A lower \( ext{pOH}\) means a more basic solution.
- A neutral solution at 25°C has a \(\text{pOH}\) of 7.
- Higher \( ext{pOH}\) values can mean increased acidity, with fewer hydroxide ions compared to hydrogen ions.
Logarithmic Functions
In pH and pOH calculations:
- The logarithm base 10 is used. This simplifies smaller concentrations into positive numbers.
- These calculations often result in straightforward whole numbers or common decimal numbers, making it easier to understand the acidity or basicity of a solution.
The premise is similar for \(\text{pOH}\), using \([\text{OH}^-]\) concentrations. If \([\text{OH}^-] = 1.0 \times 10^{-2} \, \text{M}\), the conversion to \(\text{pOH}\) results in a value of 2. Logarithms never handle negative ions directly but handle their concentration inversely through the positive logarithmic values.
Acid-Base Equilibrium
- Water is a classic example, maintaining a constant equilibrium expressed by the ion product \( K_w = [\text{H}^+] \times [\text{OH}^-] = 1.0 \times 10^{-14} \, \text{at 25°C} \).
For \( ext{pH}\) and \( ext{pOH}\) calculations, the relationship \[\text{pH} + \text{pOH} = 14\] holds at 25°C. This makes it easier to find one if you know the other. For example:
- With a \(\text{pH}\) of 5, \(\text{pOH}\) would equal 9, indicating the solution is acidic.
- Conversely, a \(\text{pOH}\) of 2 yields a \(\text{pH}\) of 12, indicating a basic solution.