chemistry
posted by Anna on .
What is the percent ionization of a 1.38 mol/L weak acid if its Ka = 2.7 x 106? Answer to two (2) decimal places.
I know what steps to use but have difficulty applying them to the quesiton
Can someone please complete the question for me since there are about 10 other questions just like this one on my practice sheet.
Step one is to use an ICE grid.
Step two is to use an ionization equation.
Step three is to use the equilibrium constant expression.
Step four is to solve the equation.
Step five is to use the quadratic.
Step six is to determine percent ionization by dividing ionized over initial concentration

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.