What is the pH of a 0.10M solution of NaF (Ka = 6.9x10^-4)?

Question

Please tell how you solve it.
Thanks

Answer 1

NaF is a salt, thus it will dissociate 100% into Na+ and F-.
Since the stoichiometry for the dissociation is 1 NaF : 1 F- you will have "initial" concentration of F-= concentration of NaF= 0.10 M

Since HF is a weak acid the ion F- will hydrolyze in water according to the reaction

.. .. .. .. .. .. .. F- +H2O <=> HF + OH-
Initial .. .. .. 0.10
React .. .. .. .x
Produce .. .. .. .. .. .. .. .. .. .. x .. .. ..x
At Equil. .. 0.10-x .. .. .. .. .. x .. .. .. x

Kb= [HF][OH-]/[F-] = x^2/(0.10-x)

Note that the concentration of H2O is considered constant in such problems and thus it is not included in the table and the K equation
We know that for the conjugate base Kb=Kw/Ka =(10^-14) / (6.9*10^-4) =1.45*10^-11

Let's assume for simplicity that 0.10 >> x so that 0.10-x=0.10
Then the equation is simplified to
x^2/0.10= 1.45*10^-11 =>
x= squareroot (0.1*1.45*10^-11 ) =1.20 *10^-6 which is << 0.10 so our assumption is valid (otherwise we would solve the quadratic to get the correct value)

pH= 14-pOH= 14-(-logx) =14-(-log(1.20*10^-6)) = 8.08

Answer 2

Lancenigo di Villorba (TV), Italy

pH = 8.1

Sodium Fluoride belongs to Weak Acid's Salts because Hydrofluoric Acid is a Weak Electrolyte.
While I dissolve Sodium Fluoride in aqueous media, it happens the following equilibrium

NaF(aq) + H2O(aq) <---> HF(aq) + NaOH(aq)

which it is able to modify the |H+| hence the pH's medium.
I remember Sodium Fluoride release Fluoride Ions obeying to Its Acid-Base's Equilibrium in aqueous medium

i) 6.9E-4 = Ka = |H+| * |F-| / |HF|

while water is a chemical compound undergoing
self-ionization

ii) 1.0E-14 = Kw = |H+| * |OH-|

On the other hand, Fluoride Mass Balancement follows

iii) 0.10 = |F-|0 = |F-| + |HF|

as it may be for Electrical Charge Balancement does

iv) |H+| + 0.10 = |H+| + |F-|0 = |H+| + |Na+| = |OH-| + |F-|

THESE FOUR EQUATIONs FORM A RESOLUBLE
MATH-SYSTEM having an interesting numerical solution.

|H+| = 8.3E-9 M hence pH = 8.1.

I hope this helps you.

Answer 3

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RE:
What is the pH of a 0.10M solution of NaF (Ka = 6.9x10^-4)?
Please tell how you solve it.
Thanks

Answer 4

- log (0.10 M) = 1