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December 21, 2014

December 21, 2014

Posted by **Nat** on Friday, July 5, 2013 at 11:52pm.

- Chemistry -
**DrBob222**, Saturday, July 6, 2013 at 12:18ampH = -log(H^+)

-3.25 = log(H^+)

Solve for (H^+). I estimated 5E-4 but you need to be more accurate than that.

..........HA ==> H^+ + A^-

I.........0.1.....0.....0

C.........-x.....x.....x

E......-0.1-x....x......x

and you know x = 5E-4 from above (again that is an estimate)

Ka = (H^+)(A^-)/(HA)

Solve for Ka.

Then .....HA ==> H^++ A^-

I........0.01.....0....0

C.........-x.....x......x

E......0.01-x.....x.....x

Substitute the E line into Ka expression along with Ka determined from the first part and solve for x = (H^+), then convert to pH.

- Chemistry -
**Nat**, Saturday, July 6, 2013 at 12:35amOh I see what I did wrong! I got the answer 3.75. Thanks for the help!

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