Posted by Cliff on Monday, December 13, 2010 at 10:45pm.
i did not exactly get that answer... but i still think this will helps. i did 10^pH formula. i got 2.818 x 10^-10 as my naswer. i toke that and divided it by the molar mass of blaech. i got the answer 3.78 x 10^12... which is simalr to your answer
I think you are trying to make this far more difficult that it is. Also, I think your prof gave you the wrong answer (or you've state something wrong or copied something wrong).
if pH = 9.55, then
pH = -log(H^+)
9.55 = -log)H^+)
-9.55 = log(H^+)
(H^+) = 2.82E-9M
(the OH^- is 3.5E-5M
(You should know your answer of pH = 13.02 isn't right. the PROBLEM states pH = 9.55 and not 13.02.
By the way, pOH = -log(OH^-), not what you wrote above.
And my problem is I can't read my calculator. For pH = 9.55, (H^+) = 2.82E-10.
I submitted an incorrect answer to that problem. The correct answer is 0.000 000 000 28 M. I apolgise.
I've been sleep deprived.
So the answer 2.8E-10 is correct. I should have rounded that last two out of the number and didn't.
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