Respuesta :
Explanation:
As it is given that the initial concentration of sulfuric acid is 0.010 M.
Hence, the ICE table for first dissociation of sulfuric acid is as follows.
[tex]H_{2}SO_{4} + H_{2}O \rightleftharpoons HSO^{-}_{4} + H_{3}O^{+}[/tex]
Initial: 0.010 0 0
Change: -0.010 +0.010 +0.010
Equilibrium: 0 0.010 0.010
When second dissociation of sulfuric acid occurs which is partial then the ICE table will be as follows.
[tex]HSO^{-}_{4} + H_{2}O \rightleftharpoons SO^{2-}_{4} + H_{3}O^{+}[/tex]
Initial: 0.010 0 0.010
Change: -x +x +x
Equilibrium: 0.010 - x x 0.010 + x
Since, it is given that [tex]K_{a} = 1.2 \times 10^{-2}[/tex]. Hence, formula for [tex]K_{a}[/tex] is as follows.
[tex]K_{a} = \frac{[SO^{2-}_{4}][H_{3}O^{+}]}{[HSO^{-}_{4}]}[/tex]
[tex]1.2 \times 10^{-2} = \frac{x(0.010 + x)}{(0.010 - x)}[/tex]
[tex]0.0012 \times (0.010 - x) = x(0.010 + x)[/tex]
[tex]0.00012 - 0.012x = x^{2} + 0.010x[/tex]
x = 0.0045
Hence, by using the equilibrium concentrations from the table and value of x we get the following.
- [tex][SO^{2-}_{4}][/tex] = x,
= 0.0045 M
- [tex][HSO^{-}_{4}][/tex] = 0.010 - x
= 0.010 - 0.0045
= 0.0055 M
- [tex][H_{3}O^{+}][/tex] = 0.010 + x
= 0.010 + 0.0045
= 0.0145 M
