Helpful Definitions

•Solutions: homogeneous mixture of two or more substances physically mixed together in a uniform way.

•Solute: substance being dissolved.

•Solvent: part of a solution doing the dissolving.

•Soluble: when a substance dissolves in another substance.

•Solubility: the ability of a substance to dissolve in another substance.

Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a solution is given by its concentration.

Molarity ( M ) = moles soluteliters of solution

Units of Molarity

0.0170 M HCl = 0.0170 moles HCl

1 L HCl solution

6.0 M HCl = 6.0 moles HCl 1 L HCl solution

Example problem #1

A solution has a volume of 250mL and contains 0.70 mol NaCl. What is its molarity?

a) .0028 M

b) 175 M

c) 2.8 M

mol

L M

Example problem #1

A solution has a volume of 250mL and contains 0.70 mol NaCl. What is its molarity?

given:

250 mL = 0.25L

mol= 0.70 mol NaCl

M = mol = 0.70 mol NaCl = 2.8M NaCl

L 0.25L

40.00 g NaOH1 mol NaOH

How many moles solute are required to make 1.35 L of 2.50 M solution?

mol = M L

What mass sodium hydroxide is this?

3.38 mol= 2.50 mol (1.35 L) L

=

3.38 mol NaOH= 135 g NaOH

Example problem #2

Example problem #3

A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution?

a) 8 M

b) 5 M

c) 2 M

Dra

no

Example problem #3

A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution?

M = 2 mole KOH = 5 M

0.4 L

Dra

no

Example problem #4NaOH is used to open stopped sinks, to treat

cellulose in the making of nylon, and to remove potato peels commercially.

If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

M = 0.20M NaOH

40.00 g NaOH

1 mol NaOH4.0 g NaOH = 0.10mol NaOH

Example problem #4

M = mol = 0.10 mol NaOH = 0.20 M NaOH L 0.500L

Example problem #5How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

a) 12 g

b) 48 g

c) 300 g

Molarity Conversion Factors

A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors.

3.0 moles NaOH and 1 L NaOH soln

1 L NaOH soln 3.0 moles NaOH

Example problem #6

Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution?

a) 15 moles HCl

b) 1.5 moles HCl

c) 0.15 moles HCl

Example problem #6

1500 mL acid soln x 1 L = 1.5 L acid soln

1000 mL

1.5 L acid soln x 0.10 mole HCl = 0.15 mole HCl

1 L acid soln

(Molarity factor)

0.342 mol

5.65 L

Example problem #7Find molarity if 58.6 g barium hydroxide are in 5.65 L solution.

171.3 g Ba(OH)2

1 mol Ba(OH)2

Step 2). What is the molarity of a 5.65 L solution containing 0.342 mol solute?

Step 1). How many moles barium hydroxide is this?

= 0.0605 M Ba(OH)2

X mol Ba(OH)2 = 58.6 g Ba(OH)2

= 0.342 mol Ba(OH)2

M =

mol

L M =

Dilution

Dilution is the process of decreasing the concentration of a stock solution by adding more solvent to the solution.

A stock solution is a concentrated solution that will be diluted to a lower concentration for actual use.

M1V1=M2V2

• M1= molarity of the stock solution• M2= molarity of the diluted solution

• V1= volume of stock solution• V2= volume of diluted solution

Dilution*notice that the number of moles are the same*but the volume of solvent has changed

M

M

M

M

M

M M

M M

M

2211 VMVM

Dilution

• Preparation of a desired solution by adding water to a concentrate.

• Moles of solute remain the same.

Dilution Example #1A stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a 100.0 mL of 0.750M?

What we know: M1 = 1.00M

V1 = ?

M2= 0.750M V2= 100.0 mL

M1V1=M2V2

75.0 mL

Dilution Example #2Concentrated HCl is 12M. What volume is needed to make 2.0L of a 1.0M solution?

What we know: M1= 12 M

V1 = ?

M2 = 1.0 M

V2 = 2.0 L

Plug in the values you have into the equation to solve for the missing value.

M1V1=M2V2

0.17 L

Dilution Example #3

• What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:

M1 = 15.8M

V1 = ?

M2 = 6.0M

V2 = 250 mL

WORK:

M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

461 g PbI2

Strategy:

1 Pb(NO3)2(aq) + 2 KI (aq) 1 PbI2(s) + 2 KNO3(aq)

89 g PbI2

1 mol PbI2

1 mol PbI2

2 mol KI

(1) Find mol KI needed to yield 89 g PbI2.(2) Based on (1), find volume of 4.0 M KI solution.

What volume of 4.0 M KI solution is required to yield 89 g PbI2?

89 g? L 4.0 M

1 L KI soln4.0 mol KI

= 0.098 L of 4.0 M KI

How many mL of a 0.500 M CuSO4 solution will react with excess Al to produce 11.0 g Cu?

__CuSO4(aq) + __Al (s)

CuSO4(aq) + Al (s) Cu(s) + Al2(SO4)3(aq)3 2 3 1? mol 11 g

__Cu(s) + __Al2(SO4)3(aq)

11 g Cu1 mol Cu

63.5 g Cu

3 mol CuSO4

3 mol Cu 0.500 mol CuSO4

1000 mL

1 L= 346 mL

1 L CuSO4 soln