A solution is prepared by placing 42.64 g of KCl in a 1.00 L
volumetric flask, adding water to dissolve the solid, and then filling
the flask to the mark. What is the molarity of an AgNO3 solution if
25.0 mL of the KCl solution react exactly with 33.2 mL of the AgNO3
solution?

convert g to moles using molecular weight of the compounds.

divide mol/ L to get Molarity  

The tricky part to the second half is that AgNO3 will react with KCl
to form silver chloride, removing some of the AgNO3.

(mols AgNO3 - mols KCl) / .0582 L

that should help

Ok I understand what you are saying but I dont understand the "(mols
AgNO3 - mols KCL)/ .0582 L" part.... I found the mols of KCL from the
given amount of grams.  But, in the balanced equation, there is a one
to one ratio so the equation would look like "(0)/.0582" and that
doesnt make sense... Can you please explain that a little bit more in
depth??  Thank you much!

I'll admit I was being lazy not calculating the mols of the two
solutions. Mea Culpa. If they are equimolar, then you are correct in
that there would be 0 mols of silver nitrate, thus 0 M.

*unless* 

you're going to use solubility constants. So, even though silver
chloride is an "insoluable" precipitate, some small amount of that
product remains in equilibrium with the products.

If you need to be that precice, you'll need to look up the Ksp of
AgCl, and calculate from there.

Hope that helps

Wow. That's heavy (no pun). I'm impressed.

tutuzdad-ga

I like to help people, but both of you (nanoalchemist and acmarcolini)
lack the basic concepts of molarity (or even common sense).
That is pathetic.  Do yourselves a favor and read your textbook and
understand the basic principles.  If you don't understand them (a
simple DEFINITION, not frigging rocket science), you don't deserve to
breathe.

Now, retards, here's the answer:

mol(AgNO3) = mol(KCl). That's trivial.

mol(AgNO3) = volume(AgNO3)*molarity(AgNO3). That's by definition.

mol(KCl) = volume(KCl)*molarity(KCl). ----------------------

Hence,
volume(AgNO3)*molarity(AgNO3) = volume(KCl)*molarity(KCl).  For retarded.

The non-trivial part (for people who smoke too much pot
to understand basic definitions, or people who can't do common sense):

(33.2ml/1000(ml/L)) * X = (25ml/1000(ml/L)) * molariry(KCl).

Notice how 1000ml/L will cancel out.  That's another common sense
thing.  If you don't understand why, see the beginning.

Now, after rearranging the equation, all stupid people will get: 
X = molarity(KCl) * 25 / 33.2.

Now, all you have to do is find the molarity of KCl (I had to add this
one after reading the answer. For the challenged).

Another definition for potheads: molarity = mol / volume.
And another for the stupid: mol = weight / Formula Weight.

Hence, molarity(KCl) = weight(KCl) / (volume*FW(KCl)).
For especially numb, molarity(KCl) = 42.64g / (1L*(39+35.5)(g/mol)).

Now, I hope this answer helps someone sober up.

Looking past acrh2?s potentially libelous imputations, let us "do
common sense," and see the whole thing written out.

The morality of the starting KCl solution is, indeed,

(1) M KCl = mols KCl / L KCl

(2) Mols KCl = grams KCl / mw KCl

	       =  42.64 g KCl/ 74.6 g/mol KCl
	       = 0.5716 mols KCl

Which, since there is 1 L of solvent (given), substituting (2) into (1) gives

0.5716 M KCl (aq)

We are also told that 25 ml of the 0.5716 M KCl solution will
?titrate? 33.2 ml of a silver nitrate solution of unknown
concentration. I read the question initially to ask ?how much silver
nitrate remains,? as would be common in a limiting reagent problem. I
address that line of questioning in greater detail below. The original
concentration of the silver nitrate solution can be determined along
acrh2?s lines:

In the 25 ml of 0.5716 M KCl solution there are 

	0.5716 mol/ L KCl   * 0.025 L KCl =
		0.01429 mols KCl

This reacts with an equimolar amount of silver nitrate, i.e.

		0.01429 mols AgNO3

These 0.01429 mols AgNO3 were in 33.2 ml solution.

We can then determine the concentration by dividing the mols by the volume

	M AgNO3 = 0.01429 mols AgNO3 / 33.2 ml 

Therefore, the starting concentration of the silver nitrate solution is:

0.4304 M AgNO3


Sample solubility problems, including AgCl can be found at
http://dl.clackamas.cc.or.us/ch105-05/solubili.htm
and
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/ksp.html
and
http://www.ncl.ac.uk/dental/oralbiol/oralenv/tutorials/ksp.htm



Whereat we find that the Ksp of Silver chloride is 1.8 E-10
The general equation is 
		Ksp = [Ag+] [Cl-]

Because the there is a one to one relationship between Ag and Cl

		Ksp=[Ag+][Ag+]
		       =[Ag+]^2

Taking the square root of both sides,

		[Ag+] = Ksp ^ (1/2)  = (1.8 E-10)^(1/2) = 1.34 E-5 M Ag+ (aq)

Which could be taken to the amount of free silver in solution, or in
the total volume of .0582 L, there are 7.8 mols of free silver able to
form silver nitrate.

nanoalchemist:
If the problem were indeed about how much free silver remained in the
solution in the form of Ag+ ions, then your reasoning is good. All the
way up to the part where you multiply 1.34e-5 M by 0.0582L and get 7.8
mol.  You must have meant to say 780 nanomol, which is about 84
micrograms of silver, which is why AgCl is insoluble enough to make
this titration quantitative at these concentrations.

The following is the best method to solve your problem.

AgNO3  +  KCl   ?   AgCl  +  KNO3     This rection takes place when
the two are reacted together. The given data is:

Mass of KCl = 42.4 g

Volume of KCl used in the reaction = V1 = 25 mL

Volume of AgNO3 used in the reaction = V2 = 33.2 mL

Molarity of KCl = not give...........we will find it out from the mass
of KCl and its volume in which the mass is dissolved.

Moles of KCl = mass in g/molar mass      ( molar mass of KCL is 74.5 g/mol)

So,         = 42.64/74.5 = 0.572 moles of KCl 
Now molarity of KCl can be calculated by applying the formula,

Molarity = no. of moles/volume of solution in litres

So.      = 0.572/1 = 0.572 mol/litre

Molarity of KCl = M1 = 0.572 mol/litre      

Molarity of AgNO3 = M2 = ?

 From the equation it is clear that 1 mole of AgNO3 reacts with 1 mole
of KCl. The following formula can now be applied to calculte the
required molarity of AgNO3.

                    M1 x V1  =  M2 x V2

                  M2 = M1 x V1/V2

Put the values.

                     = 0.572 x 25/33.2

                     = 0.43 moles/litre   is the answer