Chemistry Help

I am taking chemistry now - this semester - but I have no clue how to solve you question!!

I use Yahoo Questions/Answers for my chemistry assignments. There a lot of people willing to assist and answers any questions posted promptly.

I'll recommend it for you to try.

I am guessing because we are starting the acid & base chapter tomorrow but I think you can use the C1V1 = C2V2 equation.

So (.250M)(15.62ml) = (.100M)(V2)

V2 = 39ml

Does that work? I could be way off. Lexie

edited...

HELP!!!!!

I am having a complete brain meltdown and cannot figure out the following problem.

How many milliliters of 0.250 M phosphoric acid are required to completely neutralize 15.62 mL of a 0.100 M KOH?

Any help would be greatly appreciated.

Daryn:bluecry1:

Don't forget that each molecule of phosphoric acid aka H3(PO4) can give up 3 H+ ions to neutralize the KOH. In other words, 1/3 mole of phosphoric acid will neutralize 1 mole of KOH.

So you can figure out the number of moles of KOH you have to neutralize. (remember that the "M" stands for moles/liter or moles/1000 mL)

Then you can figure out how many ml's of .25M H3PO4 equals the number of moles of KOH.

Then divide that number of ml's by 3 since you only need 1/3 as much.

Don't forget that each molecule of phosphoric acid aka H3(PO4) can give up 3 H+ ions to neutralize the KOH. In other words, 1/3 mole of phosphoric acid will neutralize 1 mole of KOH.

So you can figure out the number of moles of KOH you have to neutralize. (remember that the "M" stands for moles/liter or moles/1000 mL)

Then you can figure out how many ml's of .25M H3PO4 equals the number of moles of KOH.

Then divide that number of ml's by 3 since you only need 1/3 as much.

thats beautiful thank you so much! I didn't even post the question but I have my last unit exam this morning and was sitting here studying I totally forgot about the H3 being worth only 1/3 as much and I solved everything else but got stuck at the end. thanks for taking the time to help.