Respuesta :
Answer;
[H+] = 0.051 M
[SO4=] = 0.051M
[HSO4-] = 0.16M
Explanation;
HSO4- <==> H+ + SO4= ..... Ka = 1.3x10^-2
Ka = [H+] [SO4=] / [HSO4-]
1.3x10^-2 = x^2 / (0.21-x)
Using algebra and solving the quadratic equation to solve for x.
x = 0.051
Therefore;
[H+] = [SO4=] = 0.051M
[HSO4-] = 0.16M
Answer : The concentration of [tex]HSO_4^-[/tex], [tex]SO_4^{2-}[/tex] and [tex]H^+[/tex] are 0.29 M, 0.061 M and 0.061 M respectively.
Explanation :
First we have to calculate the concentration of [tex]HSO_4^-[/tex]
The dissociation of [tex]KHSO_4[/tex] is:
[tex]KHSO_4\rightarrow K^++HSO_4^-[/tex]
As, 1 mole of [tex]KHSO_4[/tex] gives 1 mole of [tex]HSO_4^-[/tex]
So, 0.35 M of [tex]KHSO_4[/tex] gives 0.35 M of [tex]HSO_4^-[/tex]
Now we have to determine the concentration of [tex]SO_4^{2-}[/tex] and [tex]H^+[/tex].
The dissociation of [tex]HSO_4^-[/tex] is:
[tex]HSO_4^-\rightleftharpoons H^++SO_4^{2-}[/tex]
Initial conc. 0.35 0 0
At eqm. (0.35-x) x x
The expression of acid dissociation constant will be:
[tex]K_a=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}[/tex]
Now put all the given values in this expression, we get:
[tex]1.3\times 10^{-2}=\frac{(x)\times (x)}{(0.35-x)}[/tex]
[tex]x=0.061M[/tex]
Thus, the concentration of [tex]SO_4^{2-}[/tex] = x = 0.061 M
The concentration of [tex]H^+[/tex] = x = 0.061 M
The concentration of [tex]HSO_4^-[/tex] = 0.35 - x = 0.35 - 0.061 = 0.29 M
