[AutoGyro]

The question is:

1.   A 1000 mL bottle of anticavity mouthwash is advertised as having 0.02% sodium fluoride.
a.   Suppose that was the only dissociable ingredient in solution. What is the pH?
b.   The actual solution contains 0.02% sodium benzoate, 0.07% disodium phosphate and 0.03% sodium phosphate in addition to the sodium fluoride. What is the pH?
c.   What is the concentration of all the ions in solution?
Relevant information:
   For phosphoric acid, pKa1 = 2.12 pKa2 = 7.21 pKa3 = 12.67
   For hydrofluoric acid: pKa = 3.15
   For benzoic acid: pKa = 4.21
   Molecular weights: Sodium fluoride: 41.99 g/mol, sodium benzoate: 144.11 g/mol, sodium phosphate: 119.98 g/mol, disodium phosphate: 141.96 g/mol


My problem is with the phosphates. I know that the phosphates will control the pH of the solution because they have the higher Kb's (pKb2 = 6.79) compared to benzoate (pKb = 9.78) and fluoride (pKb = 10.85).

I have been trying for the longest time to calculate the pH and coming up with different answers at different times. I know the hydrolysis of H2A- is negligible because pKb3 is very large. I know that the dissociation of HA 2- is negligible because pKa3 is very large. That means everything is controlled by the dissociation of H2A- and hydrolysis of HA 2-. Where do I go from here?

I've been calculating the equilibrium concentration of H+ due to the dissociation of H2A- and OH- due to the hydrolysis of HA 2-. In the equilibrium equation for dissociation of H2A-, I've been putting the formal concentration of HA 2- in as the starting concentration of the dissociated HA 2- in the ICE chart of the equilibrium, and I've been doing the same thing for H2A- in the hydrolysis of HA 2- (putting its formal concentration as the initial concentration in the equilibrium). Is that correct? If so, I'm getting concentrations small enough for the auto ionization of water to come into play and getting a pH of 7.62... I don't know if that's correct.

[Borek]

I can be missing something, I am beaten - it has been long day and it is already 1 am here after clock change to winter time. My bet is: try to assume you have phosphoric buffer in the solution, then try to find out if and if - how it is disturbed by other substances.

[AutoGyro]

I can be missing something, I am beaten - it has been long day and it is already 1 am here after clock change to winter time. My bet is: try to assume you have phosphoric buffer in the solution, then try to find out if and if - how it is disturbed by other substances.

If we assume that sodium phosphate and disodium phosphate create a buffer, then:

pH = pK2 + log [HA 2-]/[H2A-]

That gives us a pH of 7.50 and [OH-] = 3.162 X 10^-7

Doesn't that mean that we should account for the self ionization of water? Do we do that by adding 1 x 10^-7 to 3.162 X 10^-7 ?

[AutoGyro]

So, treating the phosphates as a buffer, and accounting for the self ionization of water, I get a pH of 6.93 vs the pH I would get without accounting for the self ionization of water:

I used:
Kw = x(3.162 x 10^-8 + x)  where x = [OH]

To adjust from the pH I get from the HH equation.

Does this seem correct?

[AWK]

That gives us a pH of 7.50 and [OH-] = 3.162 X 10^-7

Kw = x(3.162 x 10^-8 + x)  where x = [OH]

[AutoGyro]

That gives us a pH of 7.50 and [OH-] = 3.162 X 10^-7

Kw = x(3.162 x 10^-8 + x)  where x = [OH]

pH 7.5 --> [H+] = 3.162 x 10^-8  [OH-] = 3.1622 x 10^-7

Kw = x ([H+] + x)  x = [OH-]

[Borek]

No idea why you want to account for water autoionization - just because your pH is close to 7? As long as concentration of buffer is much higher than concentration of H⁺ (have you calculated molar concentrations of phosphates?) you may safely ignore H⁺ from water.