Determine the OH- of a solution that is 0.110M in F-?

Determine the OH- of a solution that is 0.110M in F-

Answer 1

The chemical equation:

F- + H2O <==> HF + OH-

We need the Ka of HF:

6.3 x 10^-4

We need the Kb of F-:

KaKb = Kw

(6.3 x 10^-4) (Kb) = 1.0 x 10^-14

Kb = 1.5873 x 10^-11 (I won’t bother to round off yet)

Calculate [OH-]:

1.5873 x 10^-11 = [(x) (x)] / 0.110

x = 1.32 x 10^-6 M

Perhaps there was some rounding off that you did differently or perhaps the person who prepared the answer used a different Kb than you did (or was provided in the problem).

pH = 8.12

Source(s): ChemTeam

Answer 2

The correct answer is 1.7*10^-6M. First, write equilibrium F- + H20 <=> HF + H30+ entire an ICE desk. Ka = 3.5*10^-4. Nevertheless, have to use Kb worth. To seek out Kb use the connection Ka*Kb=Kw, where Kw = 1.Zero*10^-14. Kb= 2.Eighty five*10^-11. 2.Eighty five*10^-11= x^2 / 0.10-x. Assume x is smaller than zero.100 (aka small x approximation). The OH- equals 1.Sixty eight*10^-6.

Answer 3

The answer on Mastering Chemistry was 1.8*10^-6=OH-

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