Determine the pH and the concentration of all aqueous species (H^+, OH^-, H_3PO_

Question

Determine the pH and the concentration of all aqueous species (H^+, OH^-, H_3PO_4, H_2PO_4, HPO_4, and Po_4^3-) in a 0.250 M H_3P0_4 solution. Acid-Base Equilibrium and Solubility Equilibrium: Calculate the pH of the following solutions: 0.100 moles of pF and 0.120 moles NaF in 1.00 liter solution. A solution formed by mixing together equal volumes of 0.10 MNH_4NO_3 and 0.10 M NH_3. You have 0.100 moles of H_3PO_4 in a solution with a total volume of 250 mL. You then add 0.150 moles of NaOH and dilute the solution to a total volume of 1.00 L. Calculate the pH of the resulting solution, and the concentrations of all aqueous species. You are given 50.0 mL of 0.250 M citric acid (a diprotic acid with the formula H_3C_6H_5O_7; we can abbreviate lit as simply H_3C it). How many moles of sodium hydroxide would you need to add to obtain a solution with a pH of 6.70?

Explanation / Answer

pka1= 2.16 ; Ka1= 6.9*10^-3
pka2= 7.21 ;Ka2= 6,2*10^-8
pka3= 12.32Ka3= 5.0*10^-13

First we write the ICE table for the dissociation of H3PO4:

Molarity . . . . . .H3PO4 + H2O <==> H3O+ + H2PO4-
Initial . . . . . . . . .0.250 . . . . . . . . . . . . . . .0 . . . . . . .0
Change . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . . .x
Equilibrium . . . .0.250-x . . . . . . . . . . . . . . x . . . . . . .x

Now write the equilibrium constant;

Ka = [H3O+][H2PO4-] / [H3PO4]

6.9 x 10^-3 = (x)(x) / (0.250 - x)

x^2 = (6.9 x 10^-3)(0.250 - x)

x^2 = 1.7 x 10^-3 - 6.9 x 10^-3x

x^2 + 6.9 x 10^-3x - 1.7 x 10^-3 = 0

Tp solve this equation we will get

x= 0.038.

then;

[H3PO4] = 0.250 - x = 0.250 - 0.038 = 0.212 M
[H2PO4-] = x = 0.038 M = [H3O+]

Now write the ICE table for second ionization;

Molarity . . . . . . .H2PO4- + H2O <==> H3O+ + HPO4 2-
Initial . . . . . . . . . .0.038 . . . . . . . . . . . .. 0.038 . . . . . 0
Change . . . . . . . . .-x . . . . . . . . . . . . . . . . .x . . . . . . .x
Equilibrium . . . . .0.038-x . . . . . . . . . . . .0.038+x . . . . x

equilibrium constant;

Ka2 = [H3O+][HPO4 2-] / [H2PO4-]

6.2 x 10^-8 = (0.038 + x)(x) / (0.038 - x)

Here the value of Ka2 is very small thus we ignore the X then the simply above equation as follows:

(0.038)(x) / (0.038) = 6.2 x 10^-8

x = [HPO4 2-] = 6.2 x 10^-8

Now ICE table for third ionization;

Molarity . . . . . .HPO4 2- + H2O <==> H3O+ + PO4 3-
Initial . . . . . . .6.3 x 10^-8 . . . . . . . . . . . 0.038 . . . . .0
Change . . . . . . . . .-x . . . . . . . . . . . . . . .+x . . . . . .x
Equilibrium . .6.3 x 10^-8-x . .. .. . . . . .0.038+x . . . . .x


Ka3 = [H3O+][PO4 3-] / [HPO4 2-]

4.8 x 10^-13 = (0.038 + x)(x) / (6.3 x 10^-8 - x)

Here also the value of Ka3 is very small thus we ignore the X then the simply above equation as follows:

4.8 x 10^-13 = (0.038x) / (6.3 x 10^-8)
x = 8.0 x 10^-19 M = [PO4 3-]

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