1. Molarity and Dilution .

2. A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount
Solution
Solvent
Solute
Soft drink (l)
H2O
Sugar, CO2
Air (g) N2
O2, Ar, CH4
Soft Solder (s) Pb Sn
4.1

3. Solutions Solvent solute
When the solvent is water the solution is said to be aqueous

4. General Properties of Aqueous Solutions
9.1
A solution is a homogenous mixture of two or more substances.
The substance present in the largest amount (moles) is referred to as the solvent.
The other substances present are called the solutes.
A substance that dissolves in a particular solvent is said to be soluble in that solvent.

5. 9.5 Concentration of Solutions
Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.
Other common
rearrangements:

6. Worked Example 9.8
For an aqueous solution of glucose (C6H12O6), determine (a) the molarity of L of a solution that contains 50.0 g of glucose, (b) the volume of this solution that would contain mole of glucose, and (c) the number of moles of glucose in L of this solution.
Think About It Check to see that the magnitude of your answers are logical. For example, the mass given in the problem corresponds to mole of solute. If you are asked, as in part (b), for the volume that contains a number of moles smaller than 0.277, make sure your answer is smaller than the original volume.
Strategy Convert the mass of glucose given to moles, and use the equations for interconversions of M, liters, and moles to calculate the answers.
moles of glucose =
50.0 g
180.2 g/mol
= mol
Solution
molarity = = M
volume = = 1.80 L
moles of C6H12O6 in L = L×0.139 M = mol
0.277 mol C6H12O6
2.00 L solution
0.250 mol C6H12O6
0.139 M solution

7. 4.5

8. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
after dilution (f) =
MiVi
MfVf =
4.5

9. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00
Mf = 0.200
Vf = 0.06 L
Vi = ? L
Vi =
MfVf Mi =
0.200 x 0.06
4.00
= L = 3 mL
3 mL of acid
+ 57 mL of water
= 60 mL of solution

10. Another Dilution Problem
If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL
What would be the resulting concentration?
M1*V1 = M2*V2
(6.5M) * (32 mL) = M2 * (500.0 mL)
6.5 M * 32 mL
M2 =
500 mL
M2 = M

11. moles of solute before dilution = moles of solute after dilution
Concentration of Solutions
Dilution is the process of preparing a less concentrated solution from a more concentrated one.
moles of solute before dilution = moles of solute after dilution

12. (2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L)
Concentration of Solutions
In an experiment, a student needs mL of a M CuCl2 solution. A stock solution of 2.00 M CuCl2 is available.
How much of the stock solution is needed?
Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution.
(2.00 M CuCl2)(Lc) = (0.100 M CuCl2)( L)
Lc = L or 12.5 mL
To make the solution:
Pipet 12.5 mL of stock solution into a mL volumetric flask.
Carefully dilute to the calibration mark.
Mc × Lc = Md × Ld

13. Concentration of Solutions
Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as
Mc × mLc = Md × mLd

14. Worked Example 9.9
What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare mL of M HCl?
Strategy Mc = 12.0 M, Md = M, mLd = mL
Solution
12.0 M × mLc = M × mL
mLc =
0.125 M × mL
12.0 M
= 2.60 mL
Think About It Plug the answer into Equation 9.4, and make sure that the product of concentration and volume on both sides of the equation give the same result.

15. Worked Example 9.10
Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions (1 to 4) are prepared by sequential diluting mL of each solution to mL. Determine (a) the concentrations of all four standard solutions and (b) the number of moles of HCl in each solution.
Mc× mLc
mLd
Strategy (a) Md =
; (b) mol = M×L, mL = 2.500×10-1 L
Solution
(a) Md1 =
Md2 =
Md3 =
Md4 =
2.00 M × mL mL
= 8.00×10-2 M
8.00×10-2 M × mL mL
= 3.20×10-3 M
3.20×10-3 M × mL mL
= 1.28×10-4 M
1.28×10-4 M × mL mL
= 5.12×10-6 M

16. Worked Example 9.10 (cont.) Solution
(b) mol1 = 8.00×10-2 M × 2.500×10-1 L = 2.00×10-2 mol
mol2 = 3.20×10-3 M × 2.500×10-1 L = 8.00×10-4 mol
mol3 = 1.28×10-4 M × 2.500×10-1 L = 3.20×10-5 mol
mol4 = 5.12×10-6 M × 2.500×10-1 L = 1.28×10-6 mol
Think About It Serial dilution is one of the fundamental practices of homeopathy. Some remedies undergo so many serial dilutions that very few (if any) molecules of the original substance still exist in the final preparation.

17. AlCl3(s) → Al3+(aq) + 3Cl-(aq)
Worked Example 9.11
Using square-bracket notation, express the concentration of (a) chloride ion in a solution that is 1.02 M in AlCl3, (b) nitrate ion in a solution that is M in Ca(NO3)2, and (c) Na2CO3 in a solution in which [Na+] = M.
Strategy Use the concentration given in each case and the stoichiometry indicated in the corresponding chemical formula to determine the concentration of the specified ion or compound.
Solution (a) There are 3 moles of Cl- ion for every 1 mole of AlCl3,
AlCl3(s) → Al3+(aq) + 3Cl-(aq)
so the concentration of Cl- will be three times the concentration of AlCl3.
[Cl-] = [AlCl3] ×
= ×
= = 3.06 M
3 mol Cl-
1 mol AlCl3
1.02 mol AlCl3 L
3 mol Cl-
1 mol AlCl3
3.06 mol Cl- L