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