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 K
a. This constant determines the position of the equilibrium and is temperature dependant. 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 HSO
4- as opposed to the HSO
4- 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 HSO
4- ion must be greater than the others.
Further Reading
- http://en.wikipedia.org/wiki/PKa" (Wikipedia)
- http://en.wikipedia.org/wiki/Equilibrium_constant" (Wikipedia)