Question

A solution is made by combining 500 ml of 0.10 M HF with 300 ml of 0.15 M NaF. What is the pH of the solution? Ka = 7.2 × 10⁻⁴ for HF.

A) pH 2.60
B) pH 3.15
C) pH 3.40
D) pH 3.75

Answer 1

The pH of the solution is approximately 3.40

Step-by-step explanation:

To determine the pH of the solution, we need to consider the dissociation of HF in water. The Ka value of HF is given as 7.2 × 10-4. We can set up an ICE table to calculate the concentration of H3O+ and then convert it to pH.

Using the given information, we have:

1. Initial concentration of HF: 0.10 M (500 ml x 0.10 M / 800 ml = 0.0625 M)
2. Initial concentration of NaF: 0.15 M (300 ml x 0.15 M / 800 ml = 0.05625 M)
3. Change in concentration of HF and NaF: -x (since they react in a 1:1 ratio)
4. Final concentration of HF: 0.0625 M - x
5. Final concentration of NaF: 0.05625 M - x
6. Concentration of H3O+: x

After solving the equation for Ka and substituting the values, we find that x is approximately 1.61 × 10-4 M. Then, we can calculate the pH using the equation pH = -log[H3O+]. Plugging in the value for x, we find that the pH of the solution is approximately 3.40, which corresponds to option C).