Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound
ID: 941026 • 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 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.
Explanation / Answer
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