Chapter 16: Problem 60
Calculate the \(\mathrm{pH}\) of a solution that is \(0.25 \mathrm{M}\) in \(\mathrm{HF}\) and \(0.10 \mathrm{M}\) in NaF.
Short Answer
Expert verified
The pH of the solution is 2.77.
Step by step solution
01
Identify Components of the Solution
Firstly, recognize that HF is a weak acid and NaF is its salt. In this buffer solution, HF partially dissociates into \( ext{H}^+ \) and \( ext{F}^- \). NaF dissociates completely into \( ext{Na}^+ \) and \( ext{F}^- \), contributing to the \( ext{F}^- \) concentration in the solution.
02
Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for a buffer solution is \( \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \( \text{A}^- \) is \( \text{F}^- \) and \( \text{HA} \) is HF. We know \( [\text{A}^-] = 0.10 \) M and \( [\text{HA}] = 0.25 \) M.
03
Find the Acid Dissociation Constant and pKa
The acid dissociation constant \( K_a \) of HF is approximately \( 6.8 \times 10^{-4} \). Calculate the \( \text{p}K_a \) using \( \text{p}K_a = -\log(K_a) \). Therefore, \( \text{p}K_a \approx \log(6.8 \times 10^{-4}) \approx 3.17 \).
04
Substitute Values into the Equation
Substitute \( \text{p}K_a = 3.17 \), \( [\text{A}^-] = 0.10 \) M, and \( [\text{HA}] = 0.25 \) M into the Henderson-Hasselbalch equation: \( \text{pH} = 3.17 + \log \left( \frac{0.10}{0.25} \right) \).
05
Calculate the Logarithmic Term
Calculate \( \log \left( \frac{0.10}{0.25} \right) = \log(0.4) \approx -0.40 \).
06
Final Calculation of pH
Combine the values to find the pH: \( \text{pH} = 3.17 - 0.40 = 2.77 \). This is the pH of the solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch equation
When working with buffer solutions, the Henderson-Hasselbalch equation is a crucial tool. It provides a simple way to calculate the pH of a buffered solution utilizing the concentrations of an acid and its conjugate base. The formula is given by: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] Here, \([\text{A}^-]\) represents the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the undissociated weak acid. This equation is particularly useful because it turns complex ideas about acid-base chemistry into simple arithmetic, making pH calculations straightforward.
- Helps in assessing the pH by considering log ratios.
- Ideal for solutions containing a weak acid and its salt.
acid dissociation constant
The acid dissociation constant, \( K_a \), is a measure of the strength of an acid in solution. It represents how completely the acid dissociates into ions. Higher values of \( K_a \) indicate a stronger acid that dissociates more completely. For our exercise, HF (hydrofluoric acid) is considered a weak acid, meaning it only partially dissociates in water.
- HF \( \rightleftharpoons \text{H}^+ + \text{F}^- \)
- The \( K_a \) value helps in understanding this balance.
weak acid dissociation
Weak acids dissociate only partly in water. This is a key characteristic that differentiates them from strong acids, which dissociate completely. In a solution of HF and NaF, HF acts as a weak acid. Here is what the dissociation looks like: \[ \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \] The "double arrow" indicates that the dissociation is not complete and exists in equilibrium. This means that in the solution, there is a dynamic balance between the undissociated weak acid and the dissociated ions.
- Partial dissociation influences buffer capacity.
- Essential aspect for applying the Henderson-Hasselbalch equation.
pKa calculation
The \( \text{p}K_a \) value is another vital component in the world of chemistry, specifically in pH calculations. \( \text{p}K_a \) is simply the negative logarithm (\( \text{p}K_a = -\log K_a \)) of the acid dissociation constant, \( K_a \). It provides a more manageable number to convey the strength of an acid. Lower \( \text{p}K_a \) values indicate a stronger acid. For HF, the \( \text{p}K_a \) is approximately 3.17, as obtained by substituting its \( K_a \) value into the logarithmic equation.
- Simplifies comparison of acid strength.
- Essential in the Henderson-Hasselbalch equation.