[achibaby1974]

Calculate.

(a) molarity of a solution prepared by diluting 31.83 mL of 0.258 M potassium chloride to 150.00 mL
  M

ok. for this i did 0.03183 x .258 = 0.00821214    (ans)/.15 = 0.0547

 
(b) molarity of a solution prepared by diluting 25.74 mL of 0.0626 M ammonium sulfate to 500.00 mL
  M

same thing as above .2574 x 0.0626 = 0.01611324   (ans)/.5 = 0.0322


are these right? can anyone really explain what the problem means? I understand the math I just don't understand what's going on. It's a plug and chug formula but I just don't understand the concept of Molarity and its realtion to chem.  M1V1 = M2V2 M stands for mass or molarity? and what does molar mass have to do with molarity? why g per mol? basically what's is MOLARITY?? 


A question I just don't understand at all:

molarity of sodium ion in a solution made by mixing 3.51 mL of 0.299 M sodium chloride with 500. mL of 6.51  10-3 M sodium sulfate (assume volumes are additive)
                               
                            M
How do you do that? The formula? M1V1=M2V2? But then what would you be trying to find? They've given you 2 molarities and 2 volumes. What's this mean?


Last Question:
How many milliliters of 0.55 M HCl are needed to react with 6.9 g of CaCO3?
2 HCl(aq) + CaCO3(s)  CaCl2(aq) + CO2(g) + H2O(l)

I can't use the formula now right? What do you do?



Anybody know a simple way of doing this stuff? My chem prof said it was very easy and that chemist use molarity a lot b/c it saves them a lot of work and time but how really? I just don't see it.

[Arkcon]

Anybody know a simple way of doing this stuff? My chem prof said it was very easy and that chemist use molarity a lot b/c it saves them a lot of work and time but how really? I just don't see it.

Molarity questions are very common on this board.  Try to search around a little bit.  They seem to stymie a lot of people, but they are important, and pretty intuitive, once you've studied around a bit on the questions.

My standard tips:

The balanced chemical equation is an equation, it is an actual formula for converting moles of reactant into moles of product.

And yes, the numbers before the chemical formulas are not numbers of molecules, they are numbers of moles.  The units, for a balanced chemical equation, are moles.

Remember at the very beginning of chemistry class, when you did conversions of units from Imperial to metric?  That is what these questions are all about.

Simple questions are: You are given n moles of reactant, how many moles of product do you get?  These just use the balanced equation.

Tougher questions (these ones here):  You are given a volume and molarity or grams.  You convert to moles, solve for moles of product, then convert moles of product to grams, using the formula weight.

The even tougher problem (it's coming up, you'll see):  You'll be given a volume of gas, at a temperature and pressure.  You'll use the formula PV-nRT, to find n, which equals moles, use the balanced equation ... and so forth.

It all looks tough, but to your teacher, it's just units conversion.  You'll be seeing it all from that perspective, someday.

[azmanam]

a) correct.  good job.

b) you have the right numbers, but your answer is off by a factor of 10.  Check your math converting mL --> L.

Molarity is a concise, convenient way to discuss the amount of material is contained in a solution.  Here are some web pages to read for background information

http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson64.htm
http://dl.clackamas.edu/ch105-04/molarity.htm
http://dbhs.wvusd.k12.ca.us/webdocs/Solutions/Molarity.html

The equation we're using in this problem is a tool for converting between various dilutions of a solution.  Often a chemist only has one stock solution of a particular compound (1 M HCl, or 0.5 M NaCl, for example).  If the chemist needs to use that compound, but at a different concentration (or molarity) a dilution is required. 

Look at it qualitatively,  say you have some known quantity of a compound dissolved in a known quantity of solvent (like salt in water).  That known volume has a definite, known, amount of solute in it.  If you were to take cross-sections at various regions throughout the solution and physically count the number of molecules of solute, the number would be the same in every cross-section.  This is the concept of molarity.  Amount of solute (moles) in a unit volume of solvent (liters).

Now lets say you take that same solution - with the known number of solute molecules in every cross-section volume of solvent - and you add another known volume of solvent.  The volume of the solution has now increased, but what happened in each cross-section?  The total number of solute molecules has not changed, but there is now a larger volume of solvent for the solute molecules to be dispersed in.  So when you look at each cross-section again after adding more solvent, the number of molecules of solute in each cross-section is smaller than before (but still uniform throughout each cross-section).  this is the concept of dilution.  The amount of solute has not changed, but the volume of solvent has increased.  There are now fewer molecules of solute in each cross-section volume of the solution.

Thus, if as a chemist I needed 200 mL of a 0.25 M solution of HCl, but all I have in my lab is a 1 L solution of 1 M HCl, I now know how to prepare my 200 mL solution based on our discussion of dilution.  I can do some calculations to figure out how many molecules of solute (HCl)  (in moles) I'll need in my final 200 mL solution @ 0.25 M.  M = mol/L, so 0.25 M = x mol/0.2 L.  x mol = 0.25 M * 0.2 L.  I'll need 0.05 moles of HCl in the end. 

Then, I can do some more calculations to figure out how much volume of my 1 M stock solution already contains that 0.05 moles of HCl I eventually need.  M = mol/L, so 1 M = 0.05 mol/x L.  x L = 0.05 mol/1 M.  50 mL of my stock solution contains 0.05 moles of HCl. 

So I measure out 50 mL of my 1 M HCl stock solution.  Now I know I have the right number of moles of HCl, but I still need the correct volume and final molarity.  I can add my 50 mL of 1 M HCl stock solution to 150 mL of water (Always Add Acid), to arrive at a final volume of 200 mL of solution containing 0.05 moles of HCl. 

Double check my math.  M = mol/L.  M = 0.05 moles/0.2 L.  M = 0.25 mol/L.  So I finally arrive at the 200 mL of 0.25 M HCl that I needed all along.

But that was a lot of math!  The dilution equation streamlines that process into one equation.  M1V1 = M2V2 where M is molarity and V is volume.  I have a 1 M stock solution, and I need 200 mL of a 0.25 M solution.  I just don't know how much of the stock solution I should start with such that my dilution to 200 mL will produce that 0.25 M solution I need.  M1V1 = M2V2.  (1 M HCl)*(x L) = (0.25 M HCl)*(0.2 L).  x = (0.05 mol)/(1 M HCl).  x = 0.05 L.  So I need to start with 50 mL of my stock and dilute to 200 mL to arrive at my final 200 mL of a 0.25 M HCl solution.



c) First, do you know how many moles of sodium ion is produced when 1 mole of sodium chloride is dissolved in water?  How about when 1 mole of sodium sulfate is dissolved in water?  What are the balanced equations?

Next, how many moles of sodium chloride are there in 3.51 mL of a  0.299 M sodium chloride solution?  Same with sodium sulfate?  How many moles of sodium ion in each solution?  How many total moles of sodium are there?

finally, what is the total volume of the solution after mixing?  Now you have total moles and total volume, and you can solve for final molarity.



d) how many moles of calcium carbonate do you have?  how many moles of HCl will react with that many moles of calcium carbonate?  given that number of moles of HCl and the given molarity of HCl, how much volume of that HCl solution contains the requisite number of moles of HCl?

[azmanam]

Here's a hint into the reason why the M1V1=M2V2 equation works so well:  what are the units of M1V1?  What are the units of M2V2?  compare that to what I said earlier about the total physical number of solute molecules in the stock solution and in the final dilution.