Question

Phosphorous acid, H,PO,(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. PK 1.30 PK22 6.70 HO TOH Calculate the pH for each of the points in the titration of 50.0 mL of 2.4 MH,PO,(aq) with 2.4 M KOH(aq). before addition of any KOH: after addition of 25.0 mL KOH: after addition of 50.0 mL KOH: after addition of 75.0 mL KOH: after addition of 100.0 mL KOH: Question Source: MRG - General Chemistry publisher: University Science Books

Answer 1

Here, the chemical reactions representing the dissociation of phosphoric acid are,

H₃PO₃ + OH^-       $\rightarrow$         H₂PO^3- + H₂O

H₂PO₃- + OH^-     $\rightarrow$        HPO₃^2- + H₂O

Thus, there are two equivalence points.

Volume of KOH required at the first equivalence point = (50 mL x 2.4 M/ 2.4 M) = 50 mL

Volume of KOH required at the second equivalence point = 2 x 50 mL = 100 mL

a) Before the addition of any KOH

               H₃PO₃$\rightarrow$ H⁺ +   H₂PO₃^-

I                2.4--------------------------0--------- 0

C               -x---------------------------x-----------x

E              (2.4 M - x)----------------x-----------x

Ka1 = [H⁺][H₂PO₃^-] / [H₃PO₃] = x² / (2.4 - x)

5 x 10^-2 = x² / 2.4 - x

5 x 10^-2 (2.4 - x) = x²

x² + 0.05 x -

x² + 0.05 x - 0.12 = 0

On solving,

x = 0.322

[H⁺] = 0.322 M

pH = -log [H+] = -log [0.322] = 0.492

pH = 0.492

b) After the addition of 25.0 mL KOH

Moles of OH^- = Volume x concentration = 0.025 L x 2.4 M = 0.06 moles

Moles of H₃PO₃ = 0.05 L x 2.4 M = 0.12 moles

H₃PO₃ + OH^-                  $\rightarrow$                H₂PO₃^- + H₂O

(0.12 moles – 0.06 moles) 0.06 moles

Thus, [H₃PO₃] = [H₂PO₃^-]

It is the first equivalence point. At the first equivalence point, pH = pKa

Therefore,

pH = 1.30

c) After the addition of 50 mL KOH

pH = 1/2 (pKa1 + pKa2) =1/2 (1.30 + 6.70) = 4.0

pH = 4.0

d) After the addition of 75 mL KOH

It is the second equivalence point. At the second equivalence point, pH = pKa2

pH = pKa2 = 6.70

pH = 6.70

e) After the addition of 100 mL KOH

Concentration of HPO32- = 2.4 M x (50 mL/150 mL) = 0.8 M

Here, the base determines the pH

             HPO₃^2- + H₂O $\rightarrow$ H₂PO₃^- + OH^-

I              0.8 M-------------------------------0----------0

C               -x-----------------------------------x ----------x

E             0.8 –x-------------------------------x-----------x

K_b2 = x² / 0.8-x

(K_w/K_a2) = (10^-14/ 2 x 10^-7) = 5.01 x 10^-8 = x² / (0.8- x)

x² + (5.01 x 10^-8)x - 4.008 x 10^-8 = 0

x = 2.0017 x 10^-4

[OH-] = 2.0017 x 10^-4 M

pOH = -log[OH-] = -log (2.0017 x 10^-4 ) = 3.69

pOH = 3.69

pH = 14 – 3.69 = 10.30

pH = 10.30