Comparing Concentrations in a 0.10 M H2SO4 Solution

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In a 0.10 M H2SO4 solution, the concentrations of ions are not equal, with [H3O+] being the highest due to sulfuric acid's strong acidity. The concentration of [HSO4-] is greater than [SO4(-2)] because the acidity constant (Ka) for HSO4- is less than 1, indicating that the equilibrium lies to the left. As a result, more HSO4- ions are present compared to SO4(-2) ions at equilibrium. The concentration of [OH-] is the lowest among the ions due to the strong acidic nature of the solution. Overall, the discussion emphasizes the relationship between acidity constants and ion concentrations in acid-base equilibria.
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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 alot
 
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OH- is clearly in minor concentration.
H3O+ is clearly in major concentration, since H2SO4 is a very strong acid.
I think we agree here.
BaO said:
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 <--> SO42- + H3O+

So, the concentration of HSO4- is greater than the concentration of SO42-.

I hope I was helpful to you.

:approve: :-p
 
PPonte said:
Since the acidity constant of HSO4- is < 1,
i don't get this part , could you please explain it more?
 
BaO said:
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 Ka. This constant determines the position of the equilibrium and is temperature dependant. If Ka 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 Ka 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 Ka 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 Ka 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)
 
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Thank you, Hootenanny, for clarifying my explanation. I shouldn't do it so synthetic.
:approve: :-p
 
i think i get it , thanks a lot everyone :biggrin:
just one more question: how did you type Ka or H3O+ ... on computer?
 
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