Calculate the pH of a 0.746 M solution of NaF. The Ka for the weak acid HF is 6.8 × 10-4.

Answer 1

To calculate the pH of this solution, we need to consider the hydrolysis of the NaF salt.

NaF --> Na+ + F-

The F- ion will react with water to form HF and hydroxide ions:

F- + H2O <-> HF + OH-

Since HF is a weak acid, we can set up an equilibrium expression for its dissociation:

HF + H2O <-> H3O+ + a = [H3O+][F-] / [HF]

Given that Ka = 6.8 x 10^-4, we need to determine the concentration of HF at equilibrium.

Let x be the concentration of HF at equilibrium.

Using the equation for the dissociation of water:
[H3O+][OH-] = 1.0 x 10^-14

We can assume that the concentration of OH- is very low compared to the initial concentration of F-, thus [OH-] ≈ 0.

Now, we set up an ICE (Initial, Change, Equilibrium) table:

Initial:
[F-] = 0.746 M
[HF] = 0 M
[H3O+] = 0 M

Change:
[F-] = -x
[HF] = x
[H3O+] = x

Equilibrium:
[F-] = 0.746 - x
[HF] = x
[H3O+] = x

Finally, substitute these concentrations into the equilibrium expression for HF:

(6.8 x 10^-4) = x^2 / (0.746 - x)

We can make the assumption that x is much smaller than 0.746 M, so that we can simplify the equation to:

(6.8 x 10^-4) = x^2 / 0.746

Rearranging to solve for x:

x = √(0.746 * 6.8 x 10^-4) = 0.0577 M

Now, we can calculate the pH of the solution using the concentration of [H3O+] = x = 0.0577 M:

pH = -log[H3O+] = -log(0.0577) ≈ 1.24

Therefore, the pH of a 0.746 M solution of NaF is approximately 1.24.