Posted by Lena on Wednesday, July 22, 2009 at 9:49pm.
I did an experiment were I added some water and KHP into a beaker and phenophalene [sp?]. It was a titration lab. We slowly added calciuum hydroxide into the beaker and waited till the solution turned into a faint pink. The question is asking me to find the number of mols of calcium hydroxide used to neutralize the KHP. Can I just convert the number of mL of calcium hydroxide I used into grams. I was thinking that 1 mL = 1 gram. And then after just find the molar mass and divide.
Chem - DrBob222, Wednesday, July 22, 2009 at 10:24pm
You don't need to do that.
moles = M x L or
millimiles = M x mL.
Just remember that you probably will be asked for the moles KHP and to do that you must convert from moles Ca(OH)2 to moles KHP with the equation,
2KHP + Ca(OH)2 ==> 2H2O + CaK2P2
So obviously, moles KHP = 2 x moles Ca(OH)2
Chem - Lena, Wednesday, July 22, 2009 at 10:36pm
but M is in grams/mol.d When I said M I meant molar mass not molarity.
Chem - DrBob222, Wednesday, July 22, 2009 at 11:58pm
I repeat. If you want moles, it is M x L.
True, M is molarity and since M = moles/L, then M x L = moles.
I assume you have the molarity of the Ca(OH)2. And of course you may convert mL Ca(OH)2 to grams but chemicals react by moles and not grams. By the way, that CaK2P2 I wrote would better be written as K2P + CaP.
Chem - Lena, Thursday, July 23, 2009 at 12:03am
We weren't given the molarity of Ca(OH)2, that's why I don't understand how to do the question.
Chem - DrBob222, Thursday, July 23, 2009 at 12:59am
Were you given grams? Then grams/molar mass = moles Ca(OH)2. Then moles/L = M.
You don't have enough information so far to convert mL into grams. And you may not assume that 1 mL 1 gram unless that was the actual mass Ca(OH)2 used. You may have done the experiment to determine the molarity of the Ca(OH)2 in which case you would need the mass of the KHP used. If that's the case, mass KHP/molar mass KHP moles. Then 1/2 that is moles Ca(OH)2. Then M Ca(OH)2 = moles/L in the titration.
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