SOLUTION: I am currently enrolled in a General Chemistry course at a community college. I should apparently already know logarithms, but I don't. Please help! The problem is as follows...

Question 94815: I am currently enrolled in a General Chemistry course at a community college. I should apparently already know logarithms, but I don't. Please help!
The problem is as follows...
The pK_a_ of a solution is defined by the equation: pK_a_ = -logK_a_
Where: K_a_ = acid dissociation constant
Use the rules for logarithms and exponents to solve for K_a_ in terms of pK_a_.
(a.) K_a_ = _____________
(b.) If pK_a_ = 1.64 then K_a_ = ________________

I have been able to solve problem a. K_a_ = 10^-pK_a_^
However, I cannot figure out how to get the answer to problem b.
I know that the correct answer is 2.29E-2 but I can't figure out how they get that answer.
Any help would be greatly appreciated!
Thank you!
Amy M
a.mccubbin@insightbb.com
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
your solution to a. has an extra exponentiation symbol at the end, maybe this is why you can't get b.

K_a_=10^(-pK_a_) ... K_a_=10^(-1.64) ... K_a_=.0229 or 2.29E-2
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