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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