Date post: | 11-May-2015 |
Category: |
Education |
Upload: | regis-komperda |
View: | 6,358 times |
Download: | 2 times |
1
Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute
The amount of solute in a solution The amount of solute in a solution is given by its is given by its concentrationconcentration.
Molarity (M) = moles soluteliters of solution
2
1.0 L of 1.0 L of water was water was
used to used to make 1.0 L make 1.0 L of solution. of solution. Notice the Notice the water left water left
over.over.
3
PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.
PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.
Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22•6H•6H22OO
5.00 g • 1 mol
237.7 g = 0.0210 mol
0.0210 mol0.250 L
= 0.0841 M
Step 2: Step 2: Calculate MolarityCalculate Molarity
[NiClNiCl22•6 H•6 H22OO ] = 0.0841 M
4
Step 1: Step 1: Change mL to L.Change mL to L.
250 mL * 1L/1000mL = 0.250 L250 mL * 1L/1000mL = 0.250 L
Step 2: Step 2: Calculate.Calculate.
Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesMoles = (0.0500 mol/L) (0.250 L) = 0.0125 moles
Step 3: Step 3: Convert moles to grams.Convert moles to grams.
(0.0125 mol)(90.00 g/mol) = (0.0125 mol)(90.00 g/mol) = 1.13 g1.13 g
USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY
moles = M•Vmoles = M•V
What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, is
required to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?
5
Learning Check
How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?
1) 12 g
2) 48 g
3) 300 g
6
The Other Concentration The Other Concentration UnitUnit
MOLALITY, mMOLALITY, m
m of solution = mol solute
kilograms solvent
7Calculating Calculating ConcentrationsConcentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate molality ethylene glycol.O. Calculate molality ethylene glycol.
conc (molality) = 1.00 mol glycol0.250 kg H2O
= 4.00 molalconc (molality) = 1.00 mol glycol0.250 kg H2O
= 4.00 molal
8
Try this molality problem
• 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution.
m = mol solute / kg solvent
25 g NaCl 1 mol NaCl
58.5 g NaCl= 0.427 mol NaCl
Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg
0.427 mol NaCl
5 kg water= 0.0854 m salt water
9
Preparing SolutionsPreparing SolutionsPreparing SolutionsPreparing Solutions
• Weigh out a solid Weigh out a solid solute and dissolve in a solute and dissolve in a given quantity of given quantity of solvent.solvent.
• Dilute a concentrated Dilute a concentrated solution to give one solution to give one that is less that is less concentrated.concentrated.
10
DilutionsDilutions• Dilutions are especially important in the lab
• Solutions are shipped in concentrated form, but need to be diluted for use in labs
• Important: – When you dilute a solution you are NOT changing
the number of moles of solute, just volume of solvent
– The concentration of
the sample is the SAME
concentration as the
original solution
11
Solving a dilution Solving a dilution problemproblem• You start with 1 L of a 5 M solution. How would you make this into
a 2.5 M solution?
• What if you only wanted 1 L of the 2.5 M solution, and didn’t want to waste any of the 5 M solution?
– We know that any moles brought into the solution stay in the solution. So…
» How many moles do we need in our final solution?• m = M * V• m = 2.5 M * 1 L = 2.5 moles
» How what volume of stock solution contains that many moles?
• V = m/M• V = 2.5 moles /5 M = .5 L• We need to take out .5 L of stock solution then add water to
make 1 L of solution
12
MM11VV11=M=M22VV22
• This is the same process we used on the last slide, just combined into one equation.
• The first side represents the stock solution, the second side represents the diluted solution
– REMEMBER!!!! The volume you solve for is the volume of stock solution you need, this is not the overall answer to the problem!!!!
13
• Try these!
– If 250 mL of .10 M sodium chloride was diluted to a volume of 750 mL, what will the new concentration be?
– How would you make 50 mL of 3 M solution if you started with 30 M solution?
MM11VV11=M=M22VV22