Question
What is the pH of a buffer that is 0.351 M HF and 0.297 M LiF? The Ka for HF is 3.5 x 10-4.
Answers
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To find the pH of the buffer, we need to calculate the concentration of H+ ions in the solution.
The balanced chemical equation for the dissociation of HF is:
HF ⇌ H+ + F-
The Ka expression for HF is:
Ka = [H+][F-] / [HF]
Given that Ka = 3.5 x 10^-4, we can rearrange the equation to solve for [H+]:
[H+][F-] = Ka * [HF]
[H+] = (Ka * [HF]) / [F-]
First, we need to find the concentration of [F-]. In the buffer solution, LiF dissociates to form Li+ and F- ions. Since LiF is a strong electrolyte, it fully dissociates, and we can assume that the concentration of F- is equal to the concentration of LiF, which is 0.297 M.
Substituting the values into the equation:
[H+] = (3.5 x 10^-4 * 0.351) / 0.297
[H+] = 0.00041475
To find the pH, we can use the equation:
pH = -log[H+]
Substituting the value of [H+]:
pH = -log(0.00041475)
pH = 3.38
Therefore, the pH of the buffer is 3.38.
The balanced chemical equation for the dissociation of HF is:
HF ⇌ H+ + F-
The Ka expression for HF is:
Ka = [H+][F-] / [HF]
Given that Ka = 3.5 x 10^-4, we can rearrange the equation to solve for [H+]:
[H+][F-] = Ka * [HF]
[H+] = (Ka * [HF]) / [F-]
First, we need to find the concentration of [F-]. In the buffer solution, LiF dissociates to form Li+ and F- ions. Since LiF is a strong electrolyte, it fully dissociates, and we can assume that the concentration of F- is equal to the concentration of LiF, which is 0.297 M.
Substituting the values into the equation:
[H+] = (3.5 x 10^-4 * 0.351) / 0.297
[H+] = 0.00041475
To find the pH, we can use the equation:
pH = -log[H+]
Substituting the value of [H+]:
pH = -log(0.00041475)
pH = 3.38
Therefore, the pH of the buffer is 3.38.