Warm up

• Discuss at your table:

What does it mean if a solution is concentrated? What does it mean if a solution is dilute?

Bonus: Try to use some new vocabulary from last week!

Lab Debrief – use this data!

• Class Averages:

20.445.2

62.9

74.6

Excel demo: graphing

• Title• Units• Label axes• Intervals• Plot

X: temperature Y: solubility

Solubility as a function of temperature…

Solutions

solute

Solvent

When the solvent is water the solution is said to be aqueous

We need a way to quantify the Concentration of Solutions

Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.

Molarity is a unit, butalso can be used asa conversion factor

moles solutemolarity = liters solution

Example 1: Molarity formula

For an aqueous solution of glucose (C6H12O6), determine the molarity of a 2.00 L of a solution that contains 50.0 g of glucose

Example 2: Molarity formula

For an aqueous solution of glucose (C6H12O6), determine the volume of this solution that would contain 0.250 mole of glucose.

Example 3: Molarity formula

For an aqueous solution of glucose (C6H12O6) determine the number of moles of glucose in 0.500 L of this solution.

4.5

Example 1: Making a molar solution

Explain how to make 1 liter of 0.125 M KBr solution

Example 2: Making a molar solution

Explain how to make 500 milliliters of 0.375 M MgCl2 solution

Practice

• In class:

• HW: Complete lab

Warm up

• Discuss at your table:On a hot summer day you are sipping lemonade but it is far too sweet! What should you do if you want to dilute your beverage?

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

DilutionAdd Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=4.5

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

Concentration of Solutions

In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2solution. 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.

M1 V1 = M2 V2

How would you prepare 60.0 mL of 0.20 MHNO3 from a stock solution of 4.00 M HNO3?

M1V1 = M2V2

Another Dilution Problem

If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mLWhat would be the resulting concentration?

M1V1 = M2V2

Example 1: Dilution from stock solution

What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare 150.0 mL of 0.125 M HCl?

Practice

• Problems

• HW: Prelab

Worked Example 9.10

Strategy (a) Md =

Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions (1 to 4) are prepared by sequential diluting 10.00 mL of each solution to 250.00 mL. Determine (a) the concentrations of all four standard solutions and (b) the number of moles of HCl in each solution.

Solution(a) Md1 =

Md2 =

Md3 =

Md4 =

2.00 M × 10.00 mL250.00 mL

Mc× mLcmLd

; (b) mol = M×L, 250.00 mL = 2.500×10-1 L

= 8.00×10-2 M

8.00×10-2 M × 10.00 mL250.00 mL = 3.20×10-3 M

3.20×10-3 M × 10.00 mL250.00 mL = 1.28×10-4 M

1.28×10-4 M × 10.00 mL250.00 mL = 5.12×10-6 M

Worked Example 9.10 (cont.)

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.

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 molmol3 = 1.28�10-4 M� 2.500�10-1 L = 3.20�10-5 molmol4 = 5.12�10-6 M� 2.500�10-1 L = 1.28�10-6 mol

Worked Example 9.11

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.

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 0.451 M in Ca(NO3)2, and (c) Na2CO3 in a solution in which [Na+] = 0.124 M.

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 AlCl31.02 mol AlCl3

L3 mol Cl-

1 mol AlCl33.06 mol Cl-

L