Acid Ka Question?

Question

The Ka of HSO4- (aq) is 1.2 x 10^-2. Which of the following is the BEST approximation of the pH of a 0.0500 M solution of H2SO4 (aq) to 2 significant figures?

a) pH is exactly 1.3
b) pH is exactly 1.0
c) pH is exactly 1.6
d) pH is less than 1.0
e) pH is between 1.0 and 1.3

Please provide some explanations. Thanks!! :)

Answer 1

Okay I can explain the theory but since I left my scientific calculator in my dorm room and I'm at home you'll have to solve the resulting quadratic equation yourself.

H2SO4 is a very strong acid, so assume it all dissolves. We are left with .05 M HSO4- and .05 M H+.

We could just stop here and assume that since HSO4- is a weak acid, it's dissociation will have no effect on the pH and calculate pH by taking -log(.05) ....but here's how to take HSO4- into account in case it's not that weak and really will have an effect.

If we look at the dissociation of HSO4-, it'll dissociate into H+ and SO4(2-). All stoichiometric ratios are 1.

Now make a chart. Column 1 is Initial Concentration, Column 2 is Change, Column 3 is Final Concentration. Row 1 is H+, Row 2 is SO4(2-), Row 3 is HSO4-. I'm going to refer to each cell by it's column, then row. For example (1,3) is Column 1 (initial concentration) row 3 (HSO4-). So (1,3) would be the initial concentration of HSO4-.

(1,1) is 0 while (1,2) is .05, since H2SO4 dissolved completely. (1,3) is .0500.

(2,1) and (2,2) we'll call x, since this is the concentration of H+ or SO4(2-) which will be produced when the reaction reaches equilibrium. (2,3) will be -x, since that will be the concentration of HSO4- that will be lost when the reaction reaches equilibrium. (This is where the 1:1:1 stoichiometric ratios come in..see how there's no 2x or anything?)

Add columns 1 and 2 together to get column 3. (3,1) should be x, (3,2) should be .05 + x, while (3,3) should be 0.05 - x.

Now the definition of the equilibrium constant is concentrations of products over concentrations of reactants. So we will take (3,1) multiplied by (3,2) and divide by (3,3). Set this value equal to the Ka value given. 1.2 x 10^(-2) = x(.05 + x)/(0.05-x)

Solve for x and add to .05 to get the final concentration of H+. Take the negative log base 10 of this to get pH.

As I said above, the dissociation of HSO4- may have no effect whatsoever on the pH. But it's not an exceptionally weak acid, so to be sure, do all the calculation I just explained. Whew that was long. Good luck!

Answer 2

For the first dissociation event pKa=-3 so we can assume that it is 100%
This means that we get "initial" concentrations
[HSO4(-)]o =[H+]o= [H2SO4]o= 0.05 M

Now we set up the ICE table for the second dissociation

.. .. .. .. .. .. HSO4(-) <=> H+ +SO4(-2)
Initial .. .. .. .. 0.05 .. .. .. 0.05
Dissociate .. .. x
Produce .. .. .. .. .. .. .. .. .. x .. .. .. x
At equil .. .. 0.05-x .. . 0.05+x .. ..x

Ka2= [H+][SO4(-2)] / [HSO4(-2)] = (0.05+x)x/(0.05-x)=0.012 =>0.05x+x^2=-0.012x+0.0006 =>
x^2+0.062x -0.0006 =0
x1= 0.00851
x2= -0.0705 <0 rejected

Thus pH=-log[H+]= -log(0.05+x)= -log(0.05+0.00851) =1.23= 1.2 (for 2 significant figures)

So the correct answer is e