# Homework Help: Cocentration and molarity

1. Mar 9, 2008

### mmg0789

1. The problem statement, all variables and given/known data

What volume of water must be added to 45.0 mL of a 1.0 M solution of H2SO4 in order to create a 0.33 M H2SO4 solution?

vat contains 2.24 M hydrochloric acid solution. How many kg of Ca(OH)2 will be required to react completely (neutralize) 796 L of the solution?

2. Relevant equations

mv=mv

3. The attempt at a solution

for the first one, i tried using mv=mv, but noticed that it wouldnt make sense jsut plugging in the numbers directly b/c of the problem's wording

for the second one, i'm not sure how to start it

Last edited: Mar 9, 2008
2. Mar 9, 2008

### BlindSpot

In regards to your first question: what did you determine to be the final volume of the 0.33 M H2SO4 solution?

A good starting point for the second question is this: figure out how many moles of HCl you have in the vat then determine how many moles of Ca(OH)2 will be required to neutralize them.

3. Mar 9, 2008

### mmg0789

for the first one:
45*1=.33*x
x=136.4mL
ahh! i see now, subtract 45 from it

for the second one

M = moles/volume
moles = 2.24*796=1783.04mol
mols*(g/mol) = g = 1783.04/74 = 24g
dont get the right answer..not sure what i did wrong there

Last edited: Mar 9, 2008
4. Mar 9, 2008

### BlindSpot

I agree with your math used to determine the total mass of 1783.04 moles of Ca(OH)2:

moles Ca(OH)2 * g/moles = g of Ca(OH)2

but your calculations shown are off.

But this mass of base will not lead you to the right answer. Go back to the neutralization reaction (this should have been the first thing you did) and see if this gives you some ideas. I believe that the problem assumes something that may not be obvious. As a first hint, look at the acid dissociation constants (pKa values) of each proton in H2SO4. This may point you to a simplifying assumption (accurate or not) that will lead you to the given answer.

Last edited: Mar 9, 2008
5. Mar 9, 2008

### mmg0789

hmm, i'm not too sure what the acid dissociation constants are (as in: we havent studied that yet(?)) but something i noticed between what i came up with and the answer is is that 66/24 = 2.75. hopefully thats significant...not sure where that comes from though

Last edited: Mar 9, 2008
6. Mar 9, 2008

### BlindSpot

The first mistake your a making is only a computational one. Try going back to your equation mol * g/mol = g and running your numbers again. I agree that there are 1783.04 moles of HCl in the vat and that the molecular weight of Ca(OH)2 is 74 g/mol.

The second mistake will be easier to find if you write out a balanced chemical reaction for the neutralization. This will take the form of aA + bB → cS + dH20 where a, b, c, and d are integer values, A is the acid, B is the base, and S is the salt (byproduct).

7. Mar 9, 2008

### mmg0789

ah ok
first mistake was a dumb one
1783.04*74=131944.96 g Ca(OH)2

then for the second part, 131944.96 g is 131.944 kg
from the equation, 1 Ca(OH)2 : 2 HCl (...i guess i also could have done this part earlier when i had mols HCl)
132/2 = 66kg