pH of 0.30 M NaF solution (hydrolysis reaction)

[JessicaHelena]

Homework Statement

Calculate the pH of the following solutions: (caution: these are hydrolysis reactions)

a. an 0.30 M solution of NaF

Homework Equations

I think by hydrolysis reactions they mean that each of the given compounds is combined with $H_2O$.

Then for
a) NaF + HF <--> NaF + H_2O

The Attempt at a Solution

Since the molar ratios of the equation above are all 1:1:1:1, HF also has 0.30M. Then shouldn't it be $-\log_{10}0.30$?

 

[Borek]

No, your hydrolysis reaction is wrong.

Na⁺ is just a spectator, F^- is a base, it reacts with water. This is an equilibrium reaction, so it doesn't produce equivalent amount of HF. Plus, pH is -log([H⁺]), and concentration of H⁺ is not the same as the concentration of HF.

Hint: what is produced in the reaction of F^- with water? How is it related to the pH?

 

[JessicaHelena]

whoops I meant NaF + H_2O <--> NaOH + HF

 

[JessicaHelena]

So NaOH? That's a base so if I find pOH I can do 14-pOH to get pH... or is there something else?

 

[JessicaHelena]

Can I still say that there are 0.30 M of NaOH? But I'm guessing not because doing that and then doing -log and then subtracting it from 14 doesn't get me the right answer still...

 

[Borek]

First of all - the reaction you wrote is quite close to what you need but write it as net ionic reaction, as I said earlier, Na⁺ is just a spectator.

And yes, calculating pOH is the right thing to do. Can you think of a way of calculating concentration of OH^- from known initial concentration of F^-? It is not much different from calculating concentration of H⁺ from known Ka and concentration of a weak acid.

 

[epenguin]

This is just a standard type of problem of calculating the pH of a salt of a strong base and weak acid. It would be the same calculation except for the numbers for sodium acetate. For that you need to know a pK of HF. You need to write out the ionic dissociation which you haven't yet. You'd need to either (for me fastest) consider the species present in solution in significant concentration, write out the equations for mass conservation, for charge conservation, for equilibrium and solve for [H⁺] or [OH^-] using any acceptable approximations. Or go to the approximately one page in a textbook where this and similar calculations are explained.

HF can be a bit confusing because I said this is a weak acid, but you may have associations that make it a nasty strong acid. That however Is pure concentrated HF which is of different nature because of HF molecules' self associations (explained Wikipedia or other sources). Here you are in dilute enough aqueous solution and HF there is a weak acid.