Re: Application of Aqueous Equilibria

Let H2X = weak diprotic acid.

H2X = HX - + H+	for which there is an associated Ka1 = 5.90 X 10^-2

HX- = X-2 +  H+	Ka2

To calculate the molar mass we will ignore the small acidic contribution 
from the second reaction , which is a practical approximation since the 
second constant is said to be over a 1000 times weaker.

pH = 2.06, so [H+] = 10^-2.06

Using the first equation, we write an  equilibrium expression:

5.90 X 10^-2 = [HX-][H+]/[H2X]eq

But [HX-] = [H+] and at equilibrium , [H2X]eq= [H2X]original – [H+].

So, 5.90 X 10^-2 = [H+]^2/([H2X]original – [H+]).

Substituting for [H+]=10^-2.06, and solving for [H2X]original, we obtain

[H2X]original= 1.00 X10^-4moles/L

Multiplying by a volume of 0.250 L , and dividing the result into the mass 
of 0.225g, 
we get a molar mass of  90 g/mole.

Now for the second part of the question:

When the acid is fully neutralized ( the base will draw out both H+’s from 
the diprotic acid) we will be left with CaX, a non-neutral salt with a pH 
of 7.96. Note that the pOH= 14 – 7.96 = 6.04. Thus [OH-] = 10^-6.04.

When X-2 is in water:

X-2 + H2O = HX-1 + OH-1

Writing an equilibrium expression we get Kc = [HX-1] [ OH-1]/[X-2]. If we 
multiply top and bottom by [H+], we obtain:

Kc = [HX-1] [ OH-1][H+]/[X-2][H+]

Look carefully and you’ll notice that contained within the expression is 
an expression for Kw and  (1/Ka2).

Hence, Kc = Kw/Ka2 = [HX-1] [ OH-1]/[X-2], where [HX-1]= [ OH-1].
Remember [OH-] = 10^-6.04 and Kw = 1.0 X 10^-14. 
Making all these substituions and solving for Ka2 we get a value of 1.55 X 
10^-5, which as indicated in your question is less than 1000 times smaller 
than Ka1.

More questions? Email me at enricouva@hotmail.com and good luck on your 
test!