Chapter 17: Problem 83
Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.
Short Answer
Expert verified
The equation is \( \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[\text{BH}^+]}{[\text{B}]} \right) \).
Step by step solution
01
Understand the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for an acid-base buffer relates the pH of the solution to the pKa and the concentrations of the acid and its conjugate base: \[ \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where \([\text{A}^-]\) is the concentration of the base form and \([\text{HA}]\) is the concentration of the acid form.
02
Relate pOH to pKB
For a basic buffer, the pOH can be expressed in terms of the base and its conjugate acid. Derive the relation: \[ \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[\text{BH}^+]}{[\text{B}]} \right) \] where \([\text{BH}^+]\) is the concentration of the conjugate acid and \([\text{B}]\) is the concentration of the base.
03
Understand Kw Relationship
Recall the relationship between pH, pOH, and \(K_w\): \[ \text{pH} + \text{pOH} = 14 \] and \( K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \text{ at room temperature} \). This means if you know \(\text{pK}_b\), you can relate it to \(\text{pK}_a\) using: \[ \text{pK}_a + \text{pK}_b = 14 \]
04
Derive the pOH Buffer Equation
Using preciously discussed relationships, substitute \( \text{pOH} = \text{p}K_w - \text{pH} \) into the Henderson-Hasselbalch equation. Thus forming \( \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[BH^+]}{[B]} \right) \), providing a buffer equation for pOH.
05
Conclusion: Write Derived Equation
The derived equation analogous to the Henderson-Hasselbalch equation for pOH is: \[ \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[\text{BH}^+]}{[\text{B}]} \right) \] This is useful for calculating the OH given the base constant and concentration ratio.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
pOH buffering equation
When dealing with basic buffers, which consist of a weak base and its conjugate acid, we focus on the pOH concept rather than pH. It stems from a similar format to the Henderson-Hasselbalch equation but is tailored for bases. Instead of finding the pH of a solution, we calculate the pOH. The formula for the pOH buffering equation is expressed as:\[ \text{pOH} = \text{p}K_b + \log_{10} \left( \frac{[\text{BH}^+]}{[\text{B}]} \right) \]Here's what each symbol represents:
- \( \text{pOH} \) is the measure of hydroxide ion concentration.
- \( \text{p}K_b \) is the base dissociation constant, indicating how well the base captures protons.
- \([\text{BH}^+]\) is the concentration of the conjugate acid form.
- \([\text{B}]\) is the concentration of the base form.
pK_b and pK_a relationship
Understanding the relationship between \( \text{p}K_a \) and \( \text{p}K_b \) is essential when studying buffer systems. This connection helps indicate how easily a molecule can donate or accept protons under different conditions. At room temperature, there's a critical relationship given by:\[ \text{p}K_a + \text{p}K_b = 14 \]This equation stems from the property's behavior of water's ionization constant (\( K_w \)), which is equivalent to \( 10^{-14} \) at 25°C. Here's a quick breakdown:
- \( \text{p}K_a \) measures an acid's ability to donate protons.
- \( \text{p}K_b \) measures a base's ability to accept protons.
- The sum of the two equals 14 (at 25°C), providing a quick way to switch between acid and base perspectives for a given species.
buffer systems in chemistry
Buffer systems are vital in many fields, from biology to environmental science, because they resist drastic pH changes when acids or bases are introduced. A buffer system consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Here's an example of why buffers are important:
- Cellular environments critically depend on a stable pH for optimal enzyme activity.
- Our blood acts as a natural buffer, maintaining a pH close to 7.4.
- Adding hydroxide ions (OH-) will convert some of the weak acid to its conjugate base.
- Adding hydrogen ions (H+) will convert some of the conjugate base back to the weak acid.