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September 30, 2014

September 30, 2014

Posted by **bria** on Wednesday, September 20, 2006 at 1:38pm.

thanks

These problems must be reasoned through.

(1)mols KHP = mols NaOH

(2)mols KHP = mass KHP/molar mass

(3)Molarity NaOH = mols NaOH/mL NaOH

If KHP was contaminated with NaHP:

NaOH will titrate just as much H as if we had pure KHP. The molar mass of NaHP is less than that of KHP. Therefore, we weigh out KHP but it weighs less than if it were pure, which makes the mols KHP less (equation 2) and smaller mols in equation (3) makes the molarity ??. Or we can go another route. We weigh KHP, dividing by molar mass KHP in equation (2) when we should be dividing by a smaller "average" molar mass (since Na has a smaller atomic mass than K) produces fewer mols in (2) and that has ?? effect on the molarity in equation (3). I hope this helps. I find just putting each equation we use in the titration helps (even for percent problems), then start at the top and work through each equation until you get to the last one. In this case the last one was molarity of NaOH, but it could be carried further and ask for the effect on percent of an unknown acid. Then we just plug "a smaller number for molarity" into the percent equation and we have the effect on percent of the acid.

What is the H when you said "the NaOH will titrate just as much H as if we had pure KHP?"

Thanks for your help.

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