Question

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture.

pKa1 = 1.30

pKa2 = 6.70

Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.5 M H3PO3(aq) with 1.5 M KOH(aq).

A) before any addition of KOH

B) after addition of 25.0 mL of KOH

C) after addition of 50.0 mL of KOH

D) after addition of 75.0 mL of KOH

E) after addition of 100.0 mL of KOH

Answer 1

a) before any addition of KOH

0.05=(x^2) / (1.5-x)

x = [H+] = 0.3 M

pH = -log[H+]

pH=0.52

b) after addition of 25.0 mL of KOH

Mol H3PO3 in 50.0mL of 1.5M HClO solution = 50/1000*1.5 = 0.075 moles HClO
Mol KOH in 25.0ml of 1.5M solution = 25/1000*1.5 = 0.0375 mol KOH
These react to produce 0.0375 mol H3PO3 , and there is 0.075-0.0375 = 0.0375 mol unreacted HClO remaining.
Calculate the molarity of each compound in solution: Final volume = 50+25 = 75mL = 0.075L
H3PO3= 0.0375/0.075 = 0.5M
KCLO = 0.0375/0.075 = 0.5M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 1.30 + log ( 0.5/0.5)
pH = 1.30 + log 1.0
pH = 1.30 + 0.00
pH = 1.30

C) after addition of 50.0 mL of KOH

Mol H3PO3 in 50.0mL of 1.5M HClO solution = 50/1000*1.5 = 0.075 moles HClO
Mol KOH in 50.0ml of 1.5M solution = 50/1000*1.5 = 0.075 mol KOH
These react to produce 0.075 mol H3PO3 , and there is 0.075-0.075 = 0.0375 mol unreacted HClO remaining.
Calculate the molarity of each compound in solution: Final volume = 50+25 = 75mL = 0.075L
H3PO3= 0.0375/0.075 = 0.5M
KCLO = 0.0375/0.075 = 0.5M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 1.30 + log ( 0.5/0.5)
pH = 1.30 + log 1.0
pH = 1.30 + 0.00
pH = 1.30

D) after addition of 75.0 mL of KOH

Mol H3PO3 in 50.0mL of 1.5M HClO solution = 50/1000*1.5 = 0.075 moles HClO
Mol KOH in 75.0ml of 1.5M solution = 75/1000*1.5 = 0.1125 mol KOH
These react to produce 0.1125 mol H3PO3 , and there is 0.1125-0.075 = 0.0375 mol unreacted HClO remaining.
Calculate the molarity of each compound in solution: Final volume = 50+25 = 75mL = 0.075L
H3PO3= 0.0375/0.075 = 0.5M
KCLO = 0.0375/0.075 = 0.5M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 1.30 + log ( 0.5/0.5)
pH = 1.30 + log 1.0
pH = 1.30 + 0.00
pH = 1.30

E)  after addition of 75.0 mL of KOH

Mol H3PO3 in 50.0mL of 1.5M HClO solution = 50/1000*1.5 = 0.075 moles HClO
Mol KOH in 100.0ml of 1.5M solution = 100/1000*1.5 = 0.15 mol KOH
These react to produce 0.15 mol H3PO3 , and there is 0.15-0.075 = 0.075 mol unreacted HClO remaining.
Calculate the molarity of each compound in solution: Final volume = 50+25 = 75mL = 0.075L
H3PO3= 0.075/0.075 = 1.0M
KCLO = 0.075/0.075 = 1.0M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 1.30 + log ( 1.0/1.0)
pH = 1.30 + log 1.0
pH = 1.30 + 0.00
pH = 1.30