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April 17, 2014

April 17, 2014

Posted by **THOMAS** on Monday, July 12, 2010 at 7:38pm.

- chemistry -
**DrBob222**, Monday, July 12, 2010 at 8:07pmThomas, with the equation I gave you before you really don't need an ICE chart.

I think I had

Kb = (Kw/Ka) = (HC2H3O2)(OH^-)/(0.05)

1 x 10^014/1.8 x 10^-5) = 5.55 x 10^-10

so

5.55 x 10^-10 = (X^2)/0.05

Then X^2 = 5.55 x 10^-10 x 0.05

X = sqrt(5.55 x 10^-10 x 0.05)

This is the OH^-, convert to pOH, then to pH.

- chemistry -
**THOMAS**, Monday, July 12, 2010 at 9:51pmThank You, i got it. The answer was 8.7

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