Problem 136
Question
Derive an equation similar to the Henderson-Hasselbalch equation for a buffer composed of a weak base and its conjugate acid. Instead of relating pH to pKa and the relative concentrations of an acid and its conjugate base (as the Henderson-Hasselbalch equation does), the equation should relate pOH to pKb and the relative concentrations of a base and its conjugate acid.
Step-by-Step Solution
Verified Answer
The equation relating pOH to pKb and the relative concentrations of a base and its conjugate acid is pOH = pKb + log(\( \frac{[B]}{[BH^+]} \)).
1Step 1: Understand the Components of the Buffer
Identify the weak base (B) and its conjugate acid (BH+). The equilibrium for this system can be expressed as: B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)
2Step 2: Write the Equilibrium Expression
For the weak base equilibrium, write the equilibrium expression (Kb):Kb = \( \frac{[BH^+][OH^-]}{[B]} \)
3Step 3: Express pKb and pOH
Take the negative logarithm of both sides to express the equilibrium in terms of pKb and pOH:pKb = -log(Kb)pOH = -log[OH-]
4Step 4: Relate pOH to pKb and Concentrations
Substitute -log(Kb) for pKb and -log[OH-] for pOH in the equilibrium expression and solve for pOH:pOH = pKb + log(\( \frac{[B]}{[BH^+]} \))
Key Concepts
Buffer SolutionsWeak Base EquilibriumpOH and pKb Relationship
Buffer Solutions
A buffer solution is a specialized aqueous solution that resists changes in pH upon the addition of small amounts of acid or base. How you may ask? The answer lies in its composition: it contains a weak base (B) and its conjugate acid (BH+). Now, what makes buffer solutions so fascinating is their equilibrium behavior.
A buffer solution works through the reversible reaction of the weak base with water, producing its conjugate acid and hydroxide ions. When you introduce an acid into this finely balanced system, the additional hydrogen ions (H+) are consumed by the weak base, forming more of its conjugate acid. The presence of the base and its conjugate acid in significant quantities allows the solution to 'buffer' against pH changes, which is an essential property in many biochemical processes.
In a practical sense, buffer solutions are the unsung heroes in the world of chemistry, keeping the pH levels steady in everything from our blood to the lakes and rivers.
A buffer solution works through the reversible reaction of the weak base with water, producing its conjugate acid and hydroxide ions. When you introduce an acid into this finely balanced system, the additional hydrogen ions (H+) are consumed by the weak base, forming more of its conjugate acid. The presence of the base and its conjugate acid in significant quantities allows the solution to 'buffer' against pH changes, which is an essential property in many biochemical processes.
In a practical sense, buffer solutions are the unsung heroes in the world of chemistry, keeping the pH levels steady in everything from our blood to the lakes and rivers.
Weak Base Equilibrium
Understanding the equilibrium of weak bases is crucial for grasping how buffer solutions maintain pH levels. A weak base does not entirely dissociate in water but establishes an equilibrium between the base (B), its conjugate acid (BH+), and hydroxide ions (OH-).
The equilibrium constant for a weak base, denoted as Kb, showcases the propensity of the base to accept a proton in an aqueous solution. It's given by the expression: \( Kb = \frac{[BH^+][OH^-]}{[B]} \), where [BH+] is the concentration of the conjugate acid and [OH-] is the concentration of hydroxide ions.
So when dealing with buffer solutions, remember that the weak base and its equilibrium behavior are like the yin and yang of maintaining the ideal pH level.
The equilibrium constant for a weak base, denoted as Kb, showcases the propensity of the base to accept a proton in an aqueous solution. It's given by the expression: \( Kb = \frac{[BH^+][OH^-]}{[B]} \), where [BH+] is the concentration of the conjugate acid and [OH-] is the concentration of hydroxide ions.
So when dealing with buffer solutions, remember that the weak base and its equilibrium behavior are like the yin and yang of maintaining the ideal pH level.
pOH and pKb Relationship
Now, we dove into the meaty part: the pOH and pKb relationship. Why does this matter? Because it's the bridge between the equilibrium constant of a weak base and the measurable pH of the solution. Let's unfold this a bit.
The term pOH is a measure of the hydroxide ion concentration in a solution and is defined by the negative logarithm of the hydroxide ion concentration: \( pOH = -\log[OH^-] \). Similarly, pKb is derived from the base dissociation constant (Kb): \( pKb = -\log(Kb) \). Now, the elegance of this relationship comes into play when connecting these concepts to understand buffer solutions.
By rearranging the negative logarithms, we can derive a Henderson-Hasselbalch-like equation for weak bases: \( pOH = pKb + \log(\frac{[B]}{[BH^+]}) \). This equation relates the buffer's ability to maintain pH by showing how the ratio of the weak base to its conjugate acid governs the pOH of the solution. And there you have it, the pOH and pKb, like a mathematical dance, gracefully relating the concentrations to the power of hydroxide ions in the solution.
The term pOH is a measure of the hydroxide ion concentration in a solution and is defined by the negative logarithm of the hydroxide ion concentration: \( pOH = -\log[OH^-] \). Similarly, pKb is derived from the base dissociation constant (Kb): \( pKb = -\log(Kb) \). Now, the elegance of this relationship comes into play when connecting these concepts to understand buffer solutions.
By rearranging the negative logarithms, we can derive a Henderson-Hasselbalch-like equation for weak bases: \( pOH = pKb + \log(\frac{[B]}{[BH^+]}) \). This equation relates the buffer's ability to maintain pH by showing how the ratio of the weak base to its conjugate acid governs the pOH of the solution. And there you have it, the pOH and pKb, like a mathematical dance, gracefully relating the concentrations to the power of hydroxide ions in the solution.
Other exercises in this chapter
Problem 130
A 0.867-g sample of an unknown acid requires 32.2 mL of a 0.182 M barium hydroxide solution for neutralization. Assuming the acid is diprotic, calculate the mol
View solution Problem 131
A 25.0-mL volume of a sodium hydroxide solution requires 19.6mL of a 0.189 M hydrochloric acid for neutralization. A 10.0-mL volume of a phosphoric acid solutio
View solution Problem 143
What volume of 0.100 M sodium carbonate solution is required to precipitate 99% of the Mg from 1.00 L of 0.100 M magnesium nitrate solution?
View solution Problem 145
Find the pH of a solution prepared from 1.0 L of a 0.10 M solution of Ba(OH)2 and excess Zn(OH)2(s). The Ksp of Zn(OH)2 is 3 * 10-15, and the Kf of Zn(OH)4 2- i
View solution