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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: MJF on July 21, 2013, 07:15:03 PM

Title: Is this answer correct?
Post by: MJF on July 21, 2013, 07:15:03 PM

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) ::equil:: [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

Title: Re: Is this answer correct?
Post by: magician4 on July 21, 2013, 10:08:26 PM

step one is written a bit confused , as it should be -log(H⁺) = 6.20  :rarrow: 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

Title: Re: Is this answer correct?
Post by: antimatter101 on July 25, 2013, 03:53:48 AM

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

Title: Re: Is this answer correct?
Post by: eazye1334 on July 25, 2013, 07:24:45 AM

Quote from: antimatter101 on July 25, 2013, 03:53:48 AM

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.

Title: Re: Is this answer correct?
Post by: curiouscat on July 25, 2013, 08:40:39 AM

Quote from: antimatter101 on July 25, 2013, 03:53:48 AM

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

I really hope you don't mean random..  :)