Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound
ID: 954035 • Letter: P
Question
Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture.
Calculate the pH for each of the following points in the titration of 50.0 mL of a 3.0 M H3PO3(aq) with 3.0 M KOH(aq).
Phosphorous acid, HyPOs(aq), is a diprotic oxyacid that is PKa an important compound in industry and agriculture. pka2 1.30 6.70 OH Calculate the pH for each of the following points in the titration of 50.0 mL of a 3.0 M H3PO3(aq) with 3.0 M KOH(aq). Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH 11 Number (c) after addition of 50.0 mL of KOHI0 Number (d) after addition of 75.0 mL of KOH Number (e) after addition of 100.0 mL of KOHExplanation / Answer
(a): The chemical reaction for the first dissociation of H3PO3 is
------------- H3PO3 ----------------- > H2PO3-(aq) + H+(aq) ; Ka1 = 0.0501
Init.con : 3.0 M ,----------------------- 0 --------------- 0
Change: - 3.0y M ------------------ + 3.0y M, ------ + 3.0y M
Eqm.con:3.0(1 - y) M, ----------- 3.0y M, ------- 3.0y M
Ka1 = 0.0501 = (3.0y x 3.0y) / 3.0(1 - y)
=> y = 0.12115
Hence [H+] = 3.0y = 3.0 x 0.12115 = 0.3634 M
Since the second dissociation constant is very small, the H+ produced due to second dissociation can be neglected.
Hence pH = - log[H+] = - log0.3634 M = 0.440 (answer)
(b): Initial moles of 3.0 M H3PO3 = MxV = 3.0 mol/L x 0.050 L = 0.15 mol
When 25 mL ( = 0.025L) of 3.0 M KOH is added, moles of KOH added = MxV = 3.0 mol/L x 0.025 L = 0.075 mol
The balanced neutralization reaction is
------------- H3PO3 + 2 KOH --------- > K2HPO3 + 2H2O
Init. mol: 0.15 mol, 0.075 mol --------- 0 mol
change: - 0.0375, - 0.075, ------------- + 0.0375 mol
Final mol:0.1125 mol, 0 mol, ----------- 0.0375 mol
Total volume, Vt = 50 mL + 25 mL = 0.075 L
Hence concentration of H3PO3 after the reaction = 0.1125 mol / 0.075 L = 1.5 M
Now we can calculate the pH in the same way we calculated in part - (a) as
Ka1 = 0.0501 = (1.5y x 1.5y) / 1.5(1 - y)
=> y = 0.16682
Hence [H+] = 1.5y = 1.5 x 0.16682 = 0.25023 M
Hence pH = - log[H+] = - log0.25023 M = 0.602 (answer)
(c):
Initial moles of 3.0 M H3PO3 = MxV = 3.0 mol/L x 0.050 L = 0.15 mol
When 50 mL ( = 0.050L) of 3.0 M KOH is added, moles of KOH added = MxV = 3.0 mol/L x 0.050 L = 0.15 mol
The balanced neutralization reaction is
------------- H3PO3 + 2 KOH --------- > K2HPO3 + 2H2O
Init. mol: 0.15 mol, 0.15 mol --------- 0 mol
change: - 0.075 mol - 0.15 mol ------------- + 0.075 mol
Final mol:0.075 mol, 0 mol, ----------- 0.075 mol
Total volume, Vt = 50 mL + 50 mL = 0.10 L
Hence concentration of H3PO3 after the reaction = 0.075 mol / 0.10 L = 0.75 M
Now we can calculate the pH in the same way we calculated in part - (a) as
Ka1 = 0.0501 = (0.75y x 0.75y) / 0.75(1 - y)
=> y = 0.2272
Hence [H+] = 0.75y = 0.75 x 0.2272 = 0.1704 M
Hence pH = - log[H+] = - log0.1704 M = 0.769 (answer)
(d):
Initial moles of 3.0 M H3PO3 = MxV = 3.0 mol/L x 0.050 L = 0.15 mol
When 75 mL ( = 0.075L) of 3.0 M KOH is added, moles of KOH added = MxV = 3.0 mol/L x 0.075 L = 0.225 mol
The balanced neutralization reaction is
------------- H3PO3 + 2 KOH --------- > K2HPO3 + 2H2O
Init. mol: 0.15 mol, 0.225 mol --------- 0 mol
change: - 0.1125 mol - 0.225 mol ----- + 0.1125 mol
Final mol:0.0375 mol, 0 mol, ----------- 0.1125 mol
Total volume, Vt = 50 mL + 75 mL = 0.125 L
Hence concentration of H3PO3 after the reaction = 0.0375 mol / 0.125 L = 0.3 M
Now we can calculate the pH in the same way we calculated in part - (a) as
Ka1 = 0.0501 = (0.3y x 0.3y) / 0.3(1 - y)
=> y = 0.3336
Hence [H+] = 0.3y = 0.3 x 0.3336 = 0.10008 M
Hence pH = - log[H+] = - log0.10008 M = 1.00 (answer)
(e)