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 1.5 M H3PO3(aq) with 1.5 M KOH(aq).

pKa1= 1.30

pKa2= 6.70

(a) before addition of any KOH =_____number

(b) after addition of 25.0 mL of KOH = ______number

(c) after addition of 50.0 mL of KOH = ______number

(d) after addition of 75.0 mL of KOH =_______number

(e) after addition of 100.0 mL of KOH=______number

please explain how you got each! thanks so much

also this might help: For part (a), you only need to consider the first ionization because Ka1 >> Ka2. Use pKa1 to find the value of Ka1, then solve for x.

Answer 1

a) before adding base

Since pka1 >> pka2, the pH of the acid is calculated as a weak monoprotic acid, that is

pH = 1/2pKa1 -1/2 log {concentration]

= 1/2 x1.3 -1/2 log(1.5) =0.5619

b) When base is added the reaction is

H₃PO₃ + KOH -------> KH₂PO₃  + H2O

50x1.5 x2 25x1.5 0 0 initial meq

= 150 = 37.5

112.5 0 37.5 meq after reaction

56.25 37.5 mmol afterreaction

56.25/75 37.5/75 concentration after reaction

Now this comprises of a weak acid and its conjugate base, which constitutes a buffer.

pH of buffer is calculated by Hendersen equation as

pH = pKa + log([conj.base]/[acid])

= 1.3 + log (37.5/56.25) = 1.123

c) By similar argument now the concentratons of acid and conjugate base after adding 50 ml of base are

H₃PO₃ + KOH -------> KH₂PO₃  + H2O

150 75 0 0 meq before

75 0 75 meq afterreaction

37.5/100 0 75/100 molar concentration after reation

Thus pH = pka + log (base)/(acid)

= 1.3 + log (75/37.5)= 1.6020

d) after adding 75 ml of base

H₃PO₃ + KOH -------> KH₂PO₃  + H2O

150 112.5 0 0 meq before reaction

37.5 0 112.5 112.5 meq after reaction

18.75/125 0 112.5/125 molar concentration

Then pH = 1.3 + log (112.5/18.75) =2.078

e) After adding 100mL of KOH

H₃PO₃ + KOH -------> KH₂PO₃  + H2O

150 150 0 0 meq before reaction

0 0 150 150 meq after reaction

Now the mixture is not a buffer but an amphiprotic salt of a weak acid and strong base. The pH of such an amphi[rotic salt is given by

pH = 1/2(pKa1 + pKa2)

= 1/2 (1.3 +6.7) =4.0