Chapter 19: Problem 24
What is the component concentration ratio, \(\left[\operatorname{Pr}^{-}\right] /[\mathrm{HPr}],\) of a buffer that has a \(\mathrm{pH}\) of \(5.44\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HPr}=1.3 \times 10^{-5}\right) ?\)
Short Answer
Expert verified
The concentration ratio \[ [\text{Pr}^{-}] / [\text{HPr}] \] is approximately 3.55.
Step by step solution
01
Understand the Question
The question asks for the ratio of the concentration of the conjugate base \(\text{Pr}^{-}\) to the concentration of the weak acid \[\text{HPr} \]. The given information includes the \[ \text{pH} \] of the buffer and the acid dissociation constant \( K_a \) of \[ \text{HPr} \].
02
Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of the acid and its conjugate base. The equation is: \[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{Pr}^{-}]}{[\text{HPr}]} \]
03
Calculate \ \text{p}K_a \
First, calculate the \ \text{p}K_a \ from \ K_a \ using the formula: \[ \text{p}K_a = -\text{log} \( K_a \) \]Given: \[ K_a = 1.3 \times 10^{-5} \]So, \[ \text{p}K_a = -\text{log} (1.3 \times 10^{-5}) \approx 4.89 \]
04
Substitute Values into the Equation
Substitute the known values into the Henderson-Hasselbalch equation: \[ 5.44 = 4.89 + \text{log} \frac{[\text{Pr}^{-}]}{[\text{HPr}]} \]
05
Solve for the Ratio \[ \text{\frac{[\text{Pr}^{-}] }{[\text{HPr}]}} \]
Rearrange the equation to solve for the ratio: \[ 5.44 - 4.89 = \text{log} \frac{[\text{Pr}^{-}]}{[\text{HPr}]} \] \[ 0.55 = \text{log} \frac{[\text{Pr}^{-}]}{[\text{HPr}]} \] \[ \frac{[\text{Pr}^{-}]}{[\text{HPr}]} = 10^{0.55} \approx 3.55 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is an essential formula in chemistry, specifically for understanding buffer solutions. It connects the pH of a solution with the pKa (negative logarithm of the acid dissociation constant) and the concentrations of the acid and its conjugate base. The equation is:
\[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
Where:
\[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
Where:
- \text{pH\text} \text{ is the measure of the acidity of the solution.}
- \text{p}K_a\text is the logarithmic form of the acid dissociation constant (\text{K_a}).
- [\text{A}^-]\text is the concentration of the conjugate base.
- [\text{HA}]\text is the concentration of the weak acid.
acid dissociation constant
The acid dissociation constant, denoted as \( K_a \), is a quantitative measure of the strength of an acid in solution. It represents the equilibrium constant for the dissociation of a weak acid into its ion and proton. The formula for \( K_a \) is:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Where:
\( \text{p}K_a = -\text{log} \text{K}_a \).
For the exercise provided, with \( K_a = 1.3 \times 10^{-5} \), we find the pKa using the formula:
\( \text{p}K_a = -\text{log} (1.3 \times 10^{-5}) \ \text{ which is approximately } 4.89 \). This value is important to plug into the Henderson-Hasselbalch equation.
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Where:
- [\text{H}^+] is the concentration of hydrogen ions (protons).
- [\text{A}^-] is the concentration of the conjugate base.
- [\text{HA}] is the concentration of the weak acid.
\( \text{p}K_a = -\text{log} \text{K}_a \).
For the exercise provided, with \( K_a = 1.3 \times 10^{-5} \), we find the pKa using the formula:
\( \text{p}K_a = -\text{log} (1.3 \times 10^{-5}) \ \text{ which is approximately } 4.89 \). This value is important to plug into the Henderson-Hasselbalch equation.
pH calculation
Calculating the pH of a buffer solution involves using the Henderson-Hasselbalch equation. Here’s a step-by-step overview:
Given the pH and the \( K_a \), first determine the pKa value using the relation \ \( \text{p}K_a = -\text{log K}_a \).
With the pKa value and the known pH, use the Henderson-Hasselbalch equation \ \( \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \).
Rearrange the equation to solve for the concentration ratio \ \( \frac{[\text{A}^-]}{[\text{HA}]} \).
Substitute the known values into the equation. For the given exercise, the pH is 5.44 and the pKa is approximately 4.89.
Solve for the concentration ratio:\ \ \( 5.44 = 4.89 + \text{log} \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \).
Simplify the equation to find:\ \( 0.55 = \text{log} \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \).
Finally, we find the ratio by calculating the antilog (base 10): \ \( \frac{[\text{Pr}^{-}] }{[\text{HPr}] } = 10^{0.55} \) \ \approx 3.55.
Hence, the concentration ratio of \ \( \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \) in the buffer solution is approximately 3.55.
Hence, the concentration ratio of \ \( \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \) in the buffer solution is approximately 3.55.