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Topic: n/m  (Read 3724 times)

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x12179x

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n/m
« on: March 02, 2006, 07:32:07 PM »

n/m
« Last Edit: April 24, 2006, 07:59:56 PM by x12179x »

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Re:Help with Titration problem
« Reply #1 on: March 03, 2006, 03:13:50 AM »
and putting .01875 M in as Ag+, I can solve for the other concentrations.

Is what I did for part a. correct?

No. Concentration of Ag+ is much lower. Think about all the precipitate present in the solution. Where did it come from?
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