What is the pH of .0178M NaF??

Question

I am confused because it is a salt...

Answer 1

Unlike the other hydrogen halides (HCl, HBr and HI), HF is a weak acid in water (doesn't dissociate 100%) with pKa=3.15

Thus the salt will hydrolyze with F- acting as a base and leading to alkaline pH

.. .. .. .. .. .. F- +H2O <=> HF + OH-
Initial .. .. 0.0178
React .. .. ..x
Produce .. .. .. .. .. .. .. .. .. .x .. .. x
At Equil. 0.0178-x .. .. .. .. ..x .. .. x

Kb=[HF][OH-]/ [F-] =x^2/(0.0178-x)
Let's assume that x<< 0.0178. Then 0.0178-x=0.0178 and the equation is simplified to

Kb=x^2/0.0178 => x= squareroot (Kb*0.0178)
but Kb=Kw/Ka, with Kw=10^-14 and Ka=10^-pKa=10^-3.15,
so Kb= 10^-14/10^-3.15 =10^-10.85
and x = SQRT(0.0178*10^-10.85)= 5*10^-7

This means that [OH-]=x= 5*10^-7 M, which is in the same order of magnitude as [OH-] coming from pure water. This complicates the situation since we have to take the equilibrium of the self-dissociation of water into account and we cannot do more simplifications.

Let's assume we had pure water, then [OH-]initial=10^-7 and want to see how the hydrolysis of F- occurs under such conditions:

.. .. .. .. .. .. F- +H2O <=> HF + OH-
Initial .. .. 0.0178 .. .. .. .. .. .. .. .. 10^-7
React .. .. ..x
Produce .. .. .. .. .. .. .. .. .. .x .. .. x
At Equil. 0.0178-x .. .. .. .. ..x .. .. x+10^-7

Kb=[HF][OH-]/ [F-] = x(x+10^-7) / (0.0178-x)=10^-10.85 =>
x^2+ 10^-7x =0.0178*10^-10.85-10^-10.85x =>
x^2+1.00014*10^-7x -2.514*10^-13 =0
Solving the quadratic:

x=4.54*10^-7
pH=14-pOH= 14- (-log[OH-])= 14-(-log(x+10^-7)) =
=14- (-log(4.54*10^-7 + 10^-7)) =
=14-(-log(5.54*10^-7) =>

pH= 7.74

Answer 2

7.

(i believe it is neutral.)