Chapter 18: Problem 103
An aqueous solution buffered by benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right.\) \(\mathrm{COOH}\) ) and sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOHNa}\right)\) is 0.0500 \(\mathrm{M}\) in both compounds. Given that benzoic acid's \(K_{\mathrm{a}}\) equals \(6.4 \times 10^{-5},\) what is the pH of the solution?
Short Answer
Step by step solution
Determine the Relationship
Calculate pK_a
Apply the Henderson-Hasselbalch Equation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch equation
Acid-base equilibrium
In our benzoic acid and sodium benzoate buffer, the acidic form (\(\text{C}_6 \text{H}_5 \text{COOH}\)) can donate protons to convert into the conjugate base (\(\text{C}_6 \text{H}_5 \text{COO}^-\)). By understanding the fundamentals of acid-base equilibrium, we can make educated predictions about how a solution will behave under different conditions.
Benzoic acid
- It serves as an effective preservative in foods and beverages.
- Its inhibitory effect on microbial growth makes it a key ingredient in stabilizing products.
Sodium benzoate
When dissolved, sodium benzoate dissociates in water to form the benzoate ion, which can act as a conjugate base in an acid-base buffer system.
- This dissociation allows it to pair with benzoic acid to form an effective buffer.
pKa calculation
For benzoic acid, given \(K_a = 6.4 \times 10^{-5}\), the \(\text{p}K_a\) can be calculated as:\[\text{p}K_a = -\log(6.4 \times 10^{-5}) \approx 4.19\]This value gives us insight into the acidic properties of benzoic acid, further enabling precise pH calculations in buffer systems.
Aqueous solution chemistry
- In buffer solutions like those made from benzoic acid and sodium benzoate, water allows for the free interaction of molecules, maintaining the solutions' buffering capacity.
- The solvent properties of water are essential for ensuring that the equilibrium between the acid and its conjugate base is maintained.