find the pH of a 3.0 M solution of h3po4 in 1.0 L water given the ka values 7.1e-3, 6.2e-8, and 4.5e-13

Question

find the pH of a 3.0 M solution of h3po4 in 1.0 L water given the ka values 7.1e-3, 6.2e-8, and 4.5e-13

Answer 1

H₃PO₄ is a weak triprotic acid, that means in (aqueous) solution it can give up between between 0 and 3 protons (Hydrogen ions).

 

Each value of Ka must be looked at independently to determine the final concentration of H⁺ in the solution

 

Using the Henderson-Hasselbach equation we can use Ka¹ to find [H⁺] of the first ion:

 

Ka = [H⁺][A^-]/[HA] 

 

which converts into [HA] * Ka = [H⁺][A^-]

 

and because [H⁺] = [A^-] :: [HA] * Ka = [H⁺]²

 

When you plug in the numbers you find the answer for Ka¹ :: [H⁺] = (3.0M * 7.1x10^-3)^1/2 = .1459M

 

If you the same formula to find [H⁺] using Ka² you find that the concentration increases by 4.313x10^-4M (.0004313M)

 

It increases again (though by a very small amount) with Ka³ by 1.162x10^-6M (.000001162M)

 

The concentrations are additive so the final concentration of Hydrogen ions is .1463M

 

pH is defined as the negative log of the concentration of Hydrogen ions in solution so you just take the answer above and use your calculator to find the answer.

 

-log(.1463) = .8346 = pH

 

There's one final note however. All of the data provided used two digits therefore the final answer can only have two significant figures. So the proper answer would be pH = .83.

Answer 2

Alternatively, solving for x in the quadratic equation: x² + (7.1 * 10^-3)x - (3.0 * 7.1 * 10^-3),

using the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a yields x = 0.142M  (since x cannot be a negative concentration)

Answer 3

The three ionizations for H₃PO₄ are:

H₃PO₄ ==> H⁺ + H₂PO4^- Ka = 7.1x10^-3

H₂PO₄^- ==> H⁺ + HPO₄^2-  Ka = 6.2x10^-8

HPO₄^2- ==> H⁺ + PO₄^3-  Ka = 4.5x10^-13

Since the first Ka is really the only one that significantly affects the pH, the calculations can be done as follows:

H₃PO₄ ==> H⁺ + H₂PO₄^2-

Ka = 7.1x10^-3 = [H⁺][H₂PO₄^2-]/[H₃PO₄] = (x)(x)/3.0 -x and assuming x is small compared to 3.0, it can be neglected

7.1x10^-3 = x²/3.0

x² = 2.13x10^-2

x = 1.46x10^-1 = 0.146 M = [H⁺] and note that it is small (less than 5%) relative to 3.0 so assumption was valid.

pH = -log [H⁺] = -log 0.146 M

pH = 0.836 = 0.84 

 

Note: If you applied the same calculations for the other 2 ionizations for H₂PO₄^-and HPO₄^2- the contribution of H⁺ will be minimal and will not affect the pH significantly.