Posted by Anna on Friday, April 18, 2008 at 8:30pm.
Can someone please complete the question for me since there are about 10 other questions just like this one on my practice sheet.
First you write the ionization equation. For a weak acid, which I will call HA, it ionizes as follows:
HA <==> H^+ + A^-
Next you want to write the Ka expression.
Ka = (H^+)(A^-)/(HA) = 2.7 x 10^-6
Step one is to use an ICE grid.
For spacing reasons, this board doesn't handle multiple spaces at all; therefore, I will need to covert this a little differently. Basically, I do it vertically, like this.
I
C
E
Initial concn:
(HA) = 1.38 M
(H^+) = 0
(A^-) = 0
Change in concn:
(H^+) = +y
(A^-) = +y
(HA) = -y
Equilibrium concn:
(HA) = 1.38 - y
(H^+) = 0 + y = y
(A^-) = 0 + y = y
Step two is to use an ionization equation.
Step three is to use the equilibrium constant expression.
Step four is to solve the equation.
I will leave the equation for you to solve. You may want to try making the simplifying assumption that 1.38 - y = 1.38. That will avoid solving a quadratic IF it is almost true.
Step five is to use the quadratic.
Step six is to determine percent ionization by dividing ionized over initial concentration
percent ionization = (amount ionized/amount to start)*100.
The amount ionized is just (H^+) so you can use that for the numerator. The amount unionized will be 1.38. Then multiply by 100 to convert to percent.
Post your work if you get stuck.
1.38 x 100 = 1.4%
Is this correct?
No it isn't. I gave step by step instructions and worked most of it. You didn't follow any of them except to multiply by 100 to change to percent.