Respuesta :
Answer:
Explanation:
(a)
Before the addition of KOH :-
Given pKa1 of H3PO3 = 1.30
we know , pKa1 = - log10Ka1
Ka1 = 10-pKa1
Ka1 = 10-1.30
Ka1 = 0.0501
similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7
because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .
ICE table is :-
H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)
I 2.4 M 0 M 0 M
C - x + x + x
E (2.4 - x )M x M x M
x = degree of dissociation
Now expression of Ka1 is :
Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]
0.0501 = x2 / 2.4 - x
on solving for x by using quadratic formula , we have
x = 0.32
Now [ H+ ] = [ H2PO3-] = 0.32 M
pH = - log [H+]
pH = - log 0.32
pH = - ( - 0.495)
pH = 0.495
Hence pH before the addition of KOH = 0.495
(b)
After the addition of 25.0 mL of 2.4 M KOH :-
Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol
Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol
Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .
Hence pH = 1.30 M
(c)
After the addition of 50.0 mL of 2.4 M KOH :-
Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol
Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol
because Number of moles of H3PO4 = Number of moles of KOH
therefore , this point is the first equivalence point
and pH = pKa1 + pKa2 / 2
pH = 1.30 + 6.70 / 2
pH = 4.00
Hence pH = 4.00
(d)
After the addition of 75.0 mL of 2.4 M KOH :-
Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol
Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol
This is the half way of the second equivalence point , therefore pH is equal to pKa2 .
Hence pH = 6.70
