What is the pH of an aqueous solution that is 0.011 M HF (K_a = 3.5 * 10^-⁴) and 0.015 M NaF?

Question

Quizplus · Accepted Answer

First, we need to write out the dissociation equation of HF:
$$\ce{HF <=> H+ + F-}$$
The acid dissociation constant (Ka) of HF is given by:
$$K_a = \frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 3.5 	imes 10^{-4}$$
Now, let's consider what happens when we add NaF to the solution. NaF dissociates into Na+ and F-, and the F- ions can react with any H+ ions present to form HF. This is known as a buffer solution, where the HF/F- pair acts as a conjugate acid-base pair that can resist changes to the pH of the solution.

To calculate the pH of the solution, we need to compare the initial concentrations of HF and F- with the concentration of HF and F- in the buffer after the reaction has occurred. We can use an ICE (initial, change, equilibrium) table to help us with this:

|          | HF           | H+           | F-           |
| -------- | ------------ | ------------ | ------------ |
| Initial  | 0.011 M      | 0 M          | 0.015 M      |
| Change   | -x           | +x           | +x           |
| Equilibrium | 0.011 M - x  | x            | 0.015 M + x  |

We can now use the Ka expression to find x:
$$K_a = \frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = \frac{x^2}{0.011 - x} = 3.5 	imes 10^{-4}$$
Since x is small compared to the initial HF and F- concentrations, we can make the assumption that (0.015 + x) ≈ 0.015, and solve for x using the quadratic formula:
$$x = 3.2 	imes 10^{-3}$$
Now that we have found x, we can calculate the pH using the equation:
$$\mathrm{pH} = -\log_{10}[\ce{H+}] = -\log_{10}(3.2 	imes 10^{-3}) = 2.49$$

However, this is the pH of the solution before the addition of NaF. After adding NaF, we have formed a buffer solution, which means that the pH will change. To find the new pH, we need to use the Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\ce{F-}]}{[\ce{HF}]}ight) = 3.46$$

Therefore, the answer is C.

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