calculating the pH

Question

Find the pH and [OH^-] of a 3.5 x 10^-6 solution of H₃PO₄.

Answer 1

H3PO4 has 3 acid dissociation constants (Ka's) because it has 3 hydrogen atoms available to deprotonate. However, the first Ka will be the strongest, and each one that follows will be significantly weaker, so for most Gen Chem courses, only the first Ka will be used to find the pH. Ka1 = 7.5 * 10^-3. I will end the calculation after 1 deprotonation but please let me know if you would like me to continue it through all the Ka values of phosphoric acid.

This Ka matches the first deprotonation:

H3PO4 <--> H2PO4- + H+

Starting quantities:

[H3PO4] = 3.5 * 10^-6 M

[H2PO4-] = 0 M

[H+] = 0M

[H3PO4] will be decreasing as [H2PO4-] and [H+] are increasing. The amount that they increase and decrease is related to the coefficients of the balanced equation. Since all of their coefficients are 1, for every 1 mol that H3PO4 loses, H2PO4- and H+ will gain 1 mol.

Set up an ICE table to make an expression for the concentration of H+ in the solution:

H3PO4 H2PO4- H+

i 3.5*10^-6 0 0

c -x +x +x

e 3.5 *10^-6 - x x x

Set up an equilibrium equation using Ka1.

Ka1 = [products]^coefficients / [reactants]^coefficients

7.5 * 10^-3 = [x][x] / [3.5 * 10^-6 - x]

For general chemistry courses, the large difference in size (1000 times or more) between the Ka and initial concentration allows you to eliminate the x in the denominator of the equation. Rearrange the equation so that there are no longer any fractions.

Multiply both sides by [3.5 * 10^-6]

(7.5 x 10^-3) * (3.5*10^-6) = [x][x]

Multiply on the left side and simplify the right side

2.625 *10^-8 = x^2

Square root on both sides

x = 1.62*10^-4 M

From the ICE table, x represented the end concentrations of [H2PO4-] and [H+].

Use the [H+] to find the pH

pH = -log[H+] = -log[1.62*10^-4] = 3.79

Finally, use the relationship between H+ and OH- to find the hydroxide concentration.

[H+]*[OH-] = 1*10^-14

(1.62*10^-4)*[OH-] = 1*10^-14

Divide both sides by 1.62*10^-4

[OH-] = (1*10^-14)/(1.62*10^-4)

[OH-] = 6.17*10^-11 M