# Solve Ka2 of Sulfuric Acid in Aqueous Solution

• roq2

#### roq2

I'd really appreciate help :)

## Homework Statement

A 0.05 M aqueous solution of sodium hydrogen sulfate, NaHSO4, has a pH of 1.73. Calculate Ka2 for sulfuric acid. Sulfuric acid is a strong electroylyte, so you can ignore the hydrolysis of the HSO4- ion.

## Homework Equations

If I knew that I'd know the answer ;)

## The Attempt at a Solution

A 0.05 M aqueous solution of sodium hydrogen sulfate, NaHSO4, has a pH of 1.73. Calculate Ka2 for sulfuric acid. Sulfuric acid is a strong electrolyte, so you can ignore the hydrolysis of the HSO4- ion.

Sulfuric acid is H2SO4. The Ka1 equation of sulfuric acid would be H2SO4 + H2O <-> H3O + HSO4
I would of thought Ka2 would be HSO4- + H2O <--> SO4(2-) + H3O+, but it says to ignore the hydrolysis. In saying to ignore it because of being a strong electrolyte, I assume it means that H+ and SO4(2-) are the products of HSO4-. That looks the same to me as the hydrolysis, assuming the H+ eventually bonds with a hydronium ion.

So I guess the Ka2 must be HSO4- --> H+ + SO4(2-) giving [H][SO4(2-)]/[HSO4] = Ka2... but if it is a strong electrolyte it will completely go to products, so it would have no Ka, since dividing by zero would be infinity.

Being stuck there I try looking at the sodium hydrogen sulfate. The first thing that would happen would be for the Na and HSO4- to disassociate in water. It says the pH is 1.73, so the [H3O+] = 10^-1.73, and the concentration of SO4(2-) would be the same. The Ka of the HSO4- would be (10^-1.73)^2/0.05 = 0.0693. The equilibrium of HSO4- is the same thing as that as the Ka2 of sulfuric acid, but my answer is not matching the back of the book, which was 0.011.

I had one other question which is really simple and has nothing to solve.

I was reading a example given about titrations where you are titrating NH3 with HCl and it is at the equivalence point. The part of the solution I did not follow was using this formula:

NH3 + HCl --> NH4+ + CL-

What I don't understand is why the NH3 did not already partially react with water, since it was aqueous. Shouldn't it have reacted like NH3 + H2O <--> NH4+ + OH-?

I thought that since all acids and bases are in water that they have already reacted with water before you use them for anything unless it is a weak acid or base... like in a container of .1M NH3, it contains mostly water... shouldn't it react before anything is added?

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Your ionization of bisulfate is correct

The reaction is:

HSO4- <----> H+ + SO4-2 (1)

The Ka for this step is

Ka = [products]/[reactants] (2)

Originally at 0.05 M, the bisulfate (from NaHSO4) has been reduced by an amount equal to the amount of acid produced in the ionization reaction (1).

The Ka expression (2) now becomes,

Ka = [H+][SO4-2]/[HSO4-] (3)

since [H+] is equal to [SO4-2] and the starting bisulfate concentration (0.05M) has been reduced by an amount equal to the acid produced ([H+]), equation (3) becomes:

Ka = [x][x]/[0.05-x] (4)

pH = -log[H+] = 1.73 (5)

Can you do it from here?

Ooooooooh. Man, I feel dumb. I'm so rusty after spring break. Thank you very much, got it now.