Question

Calculate the pH of a 25.0 mL sample solution of phosphoric acid if its requires 20.5 mL of 0.10 M NaOH to neutralize the solution. The acid dissociation constants are listed in See Table 16.3.

Answer 1

The reaction of NaOH with phosphoric acid will be

H3PO4 + 3NaOH ---> Na3PO4 + 3H2O

So three moles of NaOH are required for neutralization of one moles of phosphoric acid

Moles of NaOH used = Molarity X volume = 0.1 M X 20.5 mL = 2.05 millimoles

so moles of phosphoric acid present = 2.05 / 3 millimoles = 0.683 millimoles

Molarity of phsophoric acid present = Moles / Volume = 0.683 / 25 = 0.0273 M

As per the dissociation constant values the solution is acidic mainly due to first dissociation as it is higher than the other two

Ka1 = 6.9 × 10⁻³

Ka2 = 6.2 × 10⁻⁸

.Ka3 = 4.88 × 10⁻¹³

            H3PO4    ---> H+ + H2PO4-

Initial     0.0273           0          0

Change    -x                +x       +x

Equilib     0.0273-x         x        x

Ka1 = [H+] [ H2PO4-] / [H3PO4]

6.9 × 10⁻³ = x^2 / 0.0273-x

0.000188 - 6.9 × 10^−3x = x^2

x^2 + 0.0069x - 0.000188 = 0

Solving for x

x = 0.010689 = [H+]

pH = -log [H+] = 1.971