Comparing Concentrations in a 0.10 M H2SO4 Solution

[BaO]

in a solution of 0.10 M H2SO4 , list these concentrations in order of the greatest to smallest: [H3O+], [OH-], [HSO4-], and [SO4(-2)]

the answer showed that none of those concentrations is equal to others, but i got [HSO4-]=[SO4-2] (eventhough I'm not quite sure)

could you help me out ?

thanks a lot

 

[PPonte]

OH^- is clearly in minor concentration.
H₃O⁺ is clearly in major concentration, since H₂SO₄ is a very strong acid.
I think we agree here.

the answer showed that none of those concentrations is equal to others, but i got [HSO4-]=[SO4-2] (eventhough I'm not quite sure)

Since the acidity constant of HSO₄^- is < 1, then the following reaction is far to the left:

HSO₄^- + H₂O <--> SO₄^2- + H₃O⁺

So, the concentration of HSO₄^- is greater than the concentration of SO₄^2-.

I hope I was helpful to you.

 

[BaO]

Since the acidity constant of HSO₄^- is < 1,

i don't get this part , could you please explain it more?

 

[Hootenanny]

i don't get this part , could you please explain it more?

In this reaction we have an equilibrium thus;

$$HSO_{4}^{-}_{(aq)} + H_{2}O_{(l)} \rightleftharpoons SO_{4}^{2-}_{(aq)} + H_{3}O^{+}_{(aq)}$$

Now any reaction at equilibrium has what is called an equilibrium constant, which in this case is the acidity constant K_a. This constant determines the position of the equilibrium and is temperature dependent. If K_a is high (in the case of a strong acid), then the acid has a strong tendency to dissociate and thus the equilibrium will lie to the right. However, in this case K_a is low (<<1 as the PPonte points out) therefore the equilibrium lies to the far left. Hence, at equilibrium the majority of the ions present will be the HSO₄^- as opposed to the HSO₄^- and the oxonium ions.

Further, the K_a is dependent on the concentrations of the products and reactants (water is ignored as it is in excess); the square brackets '[]' denote that we are considering concentrations;

$$K_{a} = \frac{\left[ H_{3}O^{+}_{(aq)} \right] \left[ SO_{4}^{2-}_{(aq)} \right]}{\left[ HSO_{4}^{-}_{(aq)} \right]}$$

Now, we can see here that if K_a is < 1, then the concentration of the HSO₄^- ion must be greater than the others.

Further Reading

• http://en.wikipedia.org/wiki/PKa" (Wikipedia)
• http://en.wikipedia.org/wiki/Equilibrium_constant" (Wikipedia)

 

[PPonte]

Thank you, Hootenanny, for clarifying my explanation. I shouldn't do it so synthetic.

 

[BaO]

i think i get it , thanks a lot everyone
just one more question: how did you type Ka or H3O+ ... on computer?