Calculate the ph of a 0.30 m naf solution. the ka value for hf is 7.2*10^-4

pH=8.32

Explanation:

The relevant equilibrium for this problem is

F⁻ + H₂O  ↔ HF + OH⁻

With a constant Kb of

Kb=

Kb=

To calculate the value of Kb we use the formula Kw=Ka*Kb, where Kw is the ionization constant of water, 1 * 10⁻¹⁴.

1 * 10⁻¹⁴ = 7.2*10⁻⁴ * Kb

Kb = 1.4 * 10⁻¹¹

So now we have

1.4 * 10⁻¹¹=

             We make the assumption that x<<<0.30 M, so we can rewrite the equation of Kb as:

1.4 * 10⁻¹¹=

So [OH⁻]=2.05*10⁻⁶

answer: i would say b.

explanation:

the answer is na (sodium)