1. A measurement of the concentration of a solution
MOLARITY
A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L
Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in mL of water.
First calculate the moles of solute:
1.5 g NaCl (1 mole NaCl) = moles of NaCl
58.45 g NaCl
Next convert mL to L: L of solution
Last, plug the appropriate values into the correct variables in the equation:
M = n / V = moles / L = mol/L

2. MOLARITY M = n / V = mol / L
How many grams of LiOH is needed to prepare mL of a 1.25 M solution?
First calculate the moles of solute needed:
M = n / V , now rearrange to solve for n: n = MV
n = (1.25 mol / L) ( L) = moles of solute needed
Next calculate the molar mass of LiOH: g/mol
Last, use diminsional analysis to solve for mass:
moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

3. MOLARITY M = n / V = mol / L
What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL?
First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute therefore:
47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = g/mL density of solute
Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass.
0.705 g/mL (1 mole/ 128 g) = mol/mL
Next convert mL to L:
mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

4. MOLARITY & DILUTION M1V1 = M2V2
The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged.
Since the amount or moles of solute before dilution (nb) and the moles of solute after the dilution (na) are the same: nb = na
And the moles for any solution can be calculated by n=MV
A relationship can be established such that
MbVb = nb = na = MaVa
Or simply : MbVb = MaVa

5. M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI
MOLARITY & DILUTION
Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar).
M1 = 0.05 mol/L M2 = ?
V1 = 25.0 mL V2 = = 75.0 mL
M1V1 = M2V2
M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = M of KI
V mL

6. M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl
MOLARITY & DILUTION
Given a 6.00 M HCl solution, how would you prepare mL of M HCl?
M1 = 6.00 mol/L M2 = 0.150
V1 = ? mL V2 = mL
M1V1 = M2V2
M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = mL of 6 M HCl
M mol/L
You would need 6.25 mL of the 6.00 M HCl reagent which would be added to about 100 mL of DI water in a mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at mL. Mix well.

7. MOLARITY & Stoichiometry
In the past chapters, you measured out a solid reactant and needed to predict the mass of the product. You learned to use the molar mass and mole ratio to achieve that goal. The same reasoning applies with solution chemistry but instead to using the molar mass you will use the molarity to convert to moles.
Stoichiometry Flow chart

8. MOLARITY & Stoichiometry
How much calcium carbonate will be precipitated by adding 25.0 mL calcium chloride to 25.0 mL of 0.56 M potassium carbonate?
CaCl K2CO3  CaCO KCl
V = 25.0 mL M = 0.56 mol/L m=?
V = 25.0 mL
First convert volume of A to moles of A:
0.025 L K2CO3 (0.56 mol/L) = moles of K2CO3
Now convert moles of A to moles of B:
0.014 mol K2CO3(1 mol CaCO3/1 mol K2CO3) =0.014 mol CaCO3
Next convert moles of B to grams of B:
0.014 mol CaCO3 (100 g/mol) = 1.40 g of CaCO3

9. MOLARITY & Stoichiometry
What would be the molarity of the potassium chloride solution from the last problem?
CaCl K2CO3  CaCO KCl
V = 25.0 mL M = 0.56 mol/L M = ?
V = 25.0 mL
First convert volume of A to moles of A:
0.025 L K2CO3 (0.56 mol/L) = moles of K2CO3
Now convert moles of A to moles of B:
0.014 mol K2CO3(2 mol KCl /1 mol K2CO3) =0.028 mol KCl
Next convert moles of B to molarity of B:
0.028 mol KCl / L = 0.56 M of KCl

10. PRACTICE PROBLEMS #24a 2.4 M 3.00 L 1.80 L 39.6 g
_________1. What is the concentration of mL of 0.60 moles of HCl?
_________ 2. What volume of M LiOH will contain 55.3 g of LiOH?
_________ 3. How many liters of water must be added to mL of 4.50 M HBr to make a solution that is M HCl?
_________ 4. How many grams of barium sulfate that will precipitate when mL of M BaCl2 and mL of 1.70 M Na2SO4 are mixed?
5. How would you prepare mL of a M ferric chloride solution if you start with crystals of FeCl3 . 6H2O?
3.00 L
1.80 L
39.6 g
Weigh out 4.59 g of the hydrated salt to a graduated cylinder with mL then add enough DI water to make 850 mL exactly.

11. PRACTICE PROBLEM #24b
1. What is the concentration of 35.0 mL of moles of KCl?
2. How many grams of KCl is needed to prepare 50.0 mL of a 0.10 M solution?
3. How many milliliters of water must be added to 30.0 mL of 9.0 M KCl to make a solution that is 0.50 M KCl?
4. How many grams of calcium carbonate will precipitate when mL of M CaCl2 and mL of 1.70 M Na2CO3 are mixed?
5. What is the concentration of the product solution (assuming the volumes are additive) when mL of M CaCl2 and mL of 1.70 M Na2CO3 are mixed?
1.59 M
0.38 g
510 mL
17.0 g
0.425 M

12. GROUP STUDY PROBLEM #24a
______1. How many grams of K3PO4 is needed to prepare mL of a M solution?
______2. What volume of M LiOH will contain 55.3 g of LiOH?
______3. How many milliliters of water must be added to mL of 3.40 M HCl to make a solution that is M HCl?
______ 4. What is the concentration of the product solution (assuming the volumes are additive) when 24.9 mL of 0.70 M BaCl2 and 45.0 mL of 0.15 M Na3PO4 are mixed?

13. GROUP STUDY PROBLEM #24b
______1. What is the concentration of mL of 0.60 moles of NaOH?
______2. What mass of MgCl2 will be required to prepare 500 mL of M solution?
______3. How many milliliters of water must be added to 50.0 mL of 1.97 M NaOH to make a solution that is M NaOH?
______ 4. How many grams of barium sulfate that will precipitate when mL of M BaCl2 and mL of 1.70 M Na2SO4 are mixed?
______ 5. What is the concentration of the product solution (assuming the volumes are additive) when mL of M BaCl2 and mL of 1.70 M Na2SO4 are mixed?