[MJF]

The pH of an acid solution is 6.20. Calculate the Ka for the acid. The initial acid concentration is 0.010 M.

Step 1. -log (6.20) = 6.3E-7

Step 2.
  HA(aq) [H]+(aq) + [A]-(aq)
I  0.010               0        0
C -6.3E-7           +6.3E-7  + 6.3E-7
E .010 - 6.3E-7        (Both will be 6.3E-7)

Step 3. Ka = (6.3E-7)^2 / (0.010 - 6.3E-7)

Ka = 4.0E-10

[magician4]

step one is written a bit confused , as it should be -log(H⁺) = 6.20  H⁺ = 6,3 E -7

... but I take it that it is that what you had in mind

though everything else looks correct, to me it seems like you miscalculated in the final step: my result would be 4.0  E -11 instead

regards

Ingo

[antimatter101]

I wonder what E stands for - I always use 10^{(random integer)} for values.

[eazye1334]

I wonder what E stands for - I always use 10^{(random integer)} for values.

4E-11 is shorthand for 4 x 10^-11, so I would assume that's what is being used here. We always used the E shorthand in our engineering courses.

[curiouscat]

I wonder what E stands for - I always use 10^{(random integer)} for values.

I really hope you don't mean random..