A 225-mg sample of a diprotic acid is dissolved in enough water to make 250. mL

Question

A 225-mg sample of a diprotic acid is dissolved in enough water to make 250. mL of solution. The pH of this solution is 2.06. A 6.9 times 10^-3 M solution of calcium hydroxide is prepared. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the second equivalence point (as determined by a pH meter) is 7.96. The first dissociation constant for the acid (K_a_1) is 5.90 times 10^-2. Assume that the volumes of the solutions are additive, that all solutions are at 25 degree C, and that K_a_1 is at least 1000 times greater than K_a_2. a. Calculate the molar mass of the acid. b. Calculate the second dissociation constant for the acid (K_a_2).

Explanation / Answer

Solution:

(a) Since the acid is a diprotic acid, let it be H2X; (2 H because the acid is diprotic in nature)
The reaction involves in the titration is as follows:

H2X + Ca(OH)2 --------> CaX + 2H2O

That is at the end point of titration: 1 M of H2X = 1 M of Ca(OH)2
Now we know, n = MxV; n= no. of moles, M= conc. in molarity and V= volume

Therefore no. of moles of Ca(OH)2 used = (6.9x10-3)V/1000 ;[divide by 1000 because V is in mL and M in moles per litre]

no. of moles of acid is the equivalent moles of Ca(OH)2 = (6.9x10-3)x250/1000 = 1.72x10-3

Therefore, 225x10-3g/(molar mass of acid) = 1.72x10-3 mol

Or, molar mass of the acid = 225x10-3/1.72x10-3 = 130.81 g/mol

(b)

H2X(aq) <--------> H+(aq) + HX–(aq); Ka1= [H+][HX-]/[H2X]

HX–(aq) <--------> H+(aq) + X2–(aq); Ka2= [H+][X2-]/[HX-]

The Ka2 expression would be: Ka2 = [H+][X2-]/[HX-]

Solving this for [HX-] = [H+][X2-]/Ka2

Now substituting this expression into the Ka1 equation for [HX-] we have,

Ka1 = [H+][HX-]/[H2X]

Ka1 = [H+] ([H+][X2-]/Ka2)/ [H2X]

Rearranging yields Ka1 x Ka2 = [H+]2 [X2-]/ [H2X]

Now we know that [X2-] must equal to [H2X], so [X2-]/[H2X] must equal 1, therefore we have

Ka1 x Ka2 = [H+][ H+] --------------> (1)

Also at first half-titration, pH = - log[H+] = 2.06

Or, [H+] = 10-2.06 = 0.0087

And second-half titration, pH = - log[H+] = 7.96

Or, [H+] = 10-7.96 = 1.09

Therefore, from equation (1) we have,

Ka2 = [H+][ H+]/Ka1

Or, Ka2 = 0.0087x1.09/5.9x10-2 = 1.6 x 10-1

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