Question

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture pka1=1.3 pka2=6.70 Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq)

a) before addition of any 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

please help? ill rate 5 stars if correct!

Answer 1

(a): For the first dissociation of H3PO3

----------------- H3PO3 ----- > H₂PO₃^-(aq) + H+(aq); Ka1 = 10^-1.3 = 5.012*10^-2

Init.conc(M): 2.7 -------------- 0 ----------------- 0

eqm.con(M): 2.7(1 - x), ----- 2.7x, ---------- 2.7x

5.012*10^-2 = 2.7x*2.7x / 2.7(1 - x)

=> x = 0.1273

Hence [H+(aq)] = 2.7x = 2.7*0.1273 = 0.3437 M

Since Ka2 is very small the amount of H+ produced due tosecond dissociation can be neglcted.

Hence total concentration of H+ is [H+(aq)] =  0.3437 M

=> pH = - log[H+(aq)] = 0.464 (answer)

(b): H3PO3 + 2KOH ---- > K2HPO3 + 2H2O

Initial moles of H3PO3 = MV = 2.7M*0.050 L = 0.135 mol

Moles of KOH added = MV = 2.7M*0.025 L = 0.0675 mol

1 mol of H3PO3 reacts with 2 moles of KOH.

Hence 0.0675 mol of KOH that will react with the moles of H3PO3 = 0.0675 mol / 2 = 0.03375 mol

Hence moles of H3PO3 remained =  0.135 mol - 0.03375 mol = 0.10125 mol

[H3PO3] = 0.10125 mol / 0.075 = 1.35 M

----------------- H3PO3 ----- > H₂PO₃^-(aq) + H+(aq); Ka1 = 10^-1.3 = 5.012*10^-2

Init.conc(M): 1.35 -------------- 0 ----------------- 0

eqm.con(M): 1.35(1 - x), ----- 1.35x, ---------- 1.35x

5.012*10^-2 = 1.35x*1.35x / 1.35(1 - x)

=> x = 0.175

[H+(aq)] = 1.35x = 1.35*0.175 = 0.23625

=> pH = - log[H+(aq)] = 0.627 (answer)

(c) Moles of KOH added = MV = 2.7M*0.050 L = 0.135 mol

moles of H3PO3 reacted = 0.135/2 = 0.0675

Hence moles of H3PO3 remained = 0.135 - 0.0675 = 0.0675 mol

total volume = 100 mL = 0.100 L

[H3PO3] = 0.0675 mol / 0.100L = 0.675 M

=> 5.012*10^-2 = 0.675x*0.675x / 0.675(1 - x)

=> x = 0.238

[H+(aq)] = 0.675x = 0.675*0.238 = 0.16065

=> pH = - log[H+(aq)] = 0.794 (answer)

(d)

Moles of KOH added = MV = 2.7M*0.075 L = 0.2025 mol

moles of H3PO3 reacted = 0.2025/2 = 0.10125 mol

Hence moles of H3PO3 remained = 0.135 - 0.10125 = 0.03375 mol

total volume = 50+75 mL = 125 mL = 0.125 L

[H3PO3] = 0.03375 mol / 0.125L = 0.27 M

=> 5.012*10^-2 = 0.27x*0.27x / 0.27(1 - x)

=> x = 0.348

[H+(aq)] = 0.27x = 0.27*0.348 = 0.09396

=> pH = - log[H+(aq)] = 1.03 (answer)

(e)

Moles of KOH added = MV = 2.7M*0.100 L = 0.27 mol

moles of H3PO3 reacted = 0.27/2 = 0.135 mol

Hence moles of H3PO3 remained = 0.135 - 0.135 = 0 mol

Hence H3PO3 is completely neutralized.

total volume = 50+100 mL = 150 mL = 0.150 L

Hence pH = (pKa1 + pKa2) / 2 = (1.3+6.70)/2 = 4 (answer)