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taskmasters
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
maacastrobr

Answer:

pH=8.32

Explanation:

The relevant equilibrium for this problem is

F⁻ + H₂O  ↔ HF + OH⁻

With a constant Kb of

Kb=[tex]\frac{[HF][OH^{-}]}{[F^{-}]}[/tex]

Kb=[tex]\frac{x*x}{0.30-x}[/tex]

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=[tex]\frac{x*x}{0.30-x}[/tex]

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

1.4 * 10⁻¹¹=[tex]\frac{x*x}{0.30}[/tex]

[tex]4.2*10^{-12}=x^{2} \\x=2.05*10^{-6}[/tex]

So [OH⁻]=2.05*10⁻⁶

  • pOH=5.68
  • pH = 14 - pOH
  • pH=8.32