Question

Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2*10^-4

Answer 1

pH=8.32

Step-by-step explanation:

The relevant equilibrium for this problem is

F⁻ + H₂O ↔ HF + OH⁻

With a constant Kb of

Kb=
`([HF][OH^(-)])/([F^(-)])`

Kb=
`(x*x)/(0.30-x)`

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=
`(x*x)/(0.30-x)`

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

1.4 * 10⁻¹¹=
`(x*x)/(0.30)`

`4.2*10^(-12)=x^(2) \\x=2.05*10^(-6)`

So [OH⁻]=2.05*10⁻⁶

• pOH=5.68

• pH = 14 - pOH

• pH=8.32

Answer 2

This problem uses the relationship between Kb and the the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:

Kb = KaKb
1.00 x 10^-14 = 7.2 x 10^-4(x)
x = 1.39 x 10^-11

We now need to calculate the [OH¯] using the Kb expression:

1.39 x 10^-11 = x^2 / (0.30 - x)

The denominator can be neglected. Thus, x is 3.73 x 10^-6.

pOH = -log 3.73 x 10^-6 = 5.43
pH = 14-5.43 = 8.57