[Nikola95]

Hey, everyone! I just can`t seem to solve this one, so any help would be appreciated.

Aqueous solution containing both HCl and H₂SO₄ was prepared, and the resulting solution had a pH value of 0.00. After the solution was diluted 100 times with water, the pH value was then 1.92. Given the information that Kb for SO₄^2- ion is 8.33*10^-13, calculate concentrations of HCl and H₂SO₄ in the initial solution.

I thought maybe in the first case HCl completely dissociates, however H₂SO₄ completely dissociates only in the first step to produce HSO₄^- ion, which gives some amount of SO₄^2-(determined by Ka and conc. of hydronium ions), and in the second case (after dilution) both HCl and H2SO4 completely dissociate, but I am not getting the correct result.

[Borek]

HSO₄^- is not completely dissociated after dilution.

Use the information given to calculate pKa2 for sulfuric acid.

[Nikola95]

Thank you, Borek. Ok, even after taking in account HSO₄^- incomplete dissociation in the second case, I`m still not getting the right result. The correct answer is:

C_(HCl)=0.6 M
C_(H2SO4)=0.4 M

Sorry to bother you, but this is really important to me, and I just can`t figure what I am doing wrong!

[Borek]

Have you checked pKa2 value?

What does it tell you about concentration of HSO₄^- at pH=0?

What does it tell you about concentration of HSO₄^- at pH=1.93?

Pretty hard to comment on your work knowing only that it yields incorrect answer. Most likely it is wrong in some way.

[Nikola95]

Yes, I have noticed that at pH=1.93, concentration of H⁺ ions is the same as Ka₂(i.e. pKa₂=c(H⁺)), so I suppose than we would have:
HSO₄^-  H⁺ + SO₄^2-
If  c(H⁺) = Ka, then: [H+]= [H+][SO42-]/[HSO4-], meaning that at equilibrium [SO42-]=[HSO4-], which should indicate that 50 % of bisulfate ions are deprotonated (am I correct about this?). But I have no idea what to do with that fact.

Here is how I tried to solve this:
1. Let [HCl] be x, and [H₂SO₄] be y

At pH=0, [H⁺]=1 M, meaning
x + y + n = 1 (1), where n is the [H⁺] which come from HSO₄^-. 
HSO₄^-  H⁺ + SO₄^2-
As H₂SO₄ first dissociates into HSO₄^-, it gives a HSO₄^- concentration of y(the same as initial H₂SO₄), so at equilibrium we have:
Ka2= [H⁺][SO₄^2-]/[HSO₄^-]= ((x+y+n)*n)/(y-n), where x+y+n equals 1 (from equation 1), after substituting known values, we get: n=y/84.33
Back to equation (1): x+y+n=1, n=y/84.33, so: x+1.011858*y=1

2. The same principle applies:

x/100 + y/100 + n/100 = 0.012
Again, after expressing n as y in the same manner as done before, solving two equations for x and y, I get negative result.

Btw, thank you for the help

[Borek]

x/100 + y/100 + n/100 = 0.012

Why n/100 and not not some new variable, say m:

x/100 + y/100 + m = 10^-1.93

What you did for pH=0 looks OK to me, but n/100 doesn't make much sense.

[Nikola95]

What do you mean? It doesn`t matter which symbol I`m using, it`s anyway gonna be expressed in terms of y, so the two equations ( of x and y sums before and after dilution) can be solved.

[Borek]

Which symbol you use doesn't matter, but dividing by 100 does. My bet is that by blindly copying previous equation and reusing the symbol you tricked yourself onto incorrect path.

Unfortunately, you don't show what you did, you just list an equation, which is partially correct and partially incorrect, and you expect me to guess where the problem lies. Sorry, I can't read your notes from here. It won't work.

[Nikola95]

You were absolutely right! I got confused there! Thank you so, so much, I have finally solved this problem! And I like the way you guys help, not just give the final solution, this is the best forum ever! Thank you a billion times!

[Borek]

Note: while your approach is OK, problem can be solved much faster using some approximation based on the pK_a2 value.

At pH=0 we are almost two units from pK_a2, which means concentration of HSO₄^- can be safely ignored, as

[tex]\frac{[HSO_4^-]}{[SO_4^{2-}]}=10^{pKa-pH}\approx 0.01[/tex]

(this is jut a rearranged acid dissociation constant formula, nothing fancier).

If so, at pH=0

10⁰ = x+y

(where x and y have the same meaning they had in your approach - concentrations of both acids).

At pH=pK_a2 we know [HSO₄^-]=[SO₄^2-], so

10^-1.93 = x/100 + 1.5y/100

so basically we have a set of trivial equations

x+y=1
x/100+1.5y/100=10^-1.93

http://www.wolframalpha.com/input/?i=x%2By=1+and+x/100%2B1.5*y/100=10^-1.93&dataset=

x=0.65
y=0.35

which fits the data better than 0.6 & 0.4 given as the correct answer.

[Nikola95]

Interesting! Thanks again!