Concentrations of Solutions

Concentrations of Solutions The properties and behavior of solutions

often depend not only on the type of solute but also on the concentration of the solute.

Concentration: the amount of solute dissolved in a given quantity of solvent or solutionmany different concentration units

(%, ppm, g/L, etc)

often expressed as molarity

Concentrations of Solutions Molarity (M): the number of moles of solute

per liter of solution

Molarity = moles of solutevolume of solution in liters

6.0 M HCl 6.0 moles of HCl per liter of solution.

9.0 M HCl 9.0 moles of HCl per liter of solution.

Concentration of Solutions To calculate the molarity of a solution:

Divide amount of solute by volume of solutionconvert amount of solute to molesconvert volume to liters

Units of molarity are mol/L (M)

Concentration of Solutions

Example: What is the molarity of a solution prepared by dissolving 225 g of glucose (C6H12O6) in enough water to make 825 mL of solution?

M = 225 g glucose x 1000 mL x 1 mol 825 mL soln 1 L 180.2 g

M = 1.51 M glucose

Concentration of SolutionsInterconverting Molarity, Moles, and Volume

Molarity can be used as a conversion factor.

The definition of molarity contains 3 quantities:

Molarity = moles of solutevolume of solution in liters

If you know two of these quantities, you can find the third.

Concentration of Solutions Interconverting Molarity, Moles, and Volume

Example: How many moles of HCl are present in 2.5 L of 0.10 M HCl?

Given: 2.5 L of soln0.10M HCl

Find: mol HCl

Use molarity as a conversion factor

Mol HCl = 2.5 L soln x 0.10 mol HCl1 L of soln

= 0.25 mol HCl


Example: What volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH?Given: 0.50 mol NaOH

0.10 M NaOHFind: vol soln

Use M as a conversion factor

Vol soln = 0.50 mol NaOH x 1 L soln0.10 mol NaOH

= 5.0 L solution


Example: How many grams of NaNO3 are needed to prepare 250.0 mL of 1.00 M NaNO3?

Given: 250.0 mL soln 1.00 M NaNO3

Find: g NaNO3 Conversion factors:Molarity, molar mass

Strategy: mL L mol grams


g NaNO3 = 250.0 mL soln x 1 L x 1.00 mol

1000 mL 1 L soln

x 85.0 g NaNO3

1 mol

= 21.25 g NaNO3 = 21.3 g NaNO3

Concentration of Solutions Dilution Calculations

Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions).12 M HCl12 M H2SO4

More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.


A given volume of a stock solution contains a specific number of moles of solute.25 mL of 6.0 M HCl contains 0.15 mol HCl

If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change.Still contains 0.15 mol HCl


Moles solute = moles solutebefore dilution after dilution

Although the number of moles of solute does not change, the volume of solution does change.

The concentration of the solution will change since Molarity = mol soluteVol soln


When a solution is diluted, the concentration of the new solution can be found using:

Mi x Vi = Mf x Vf

where Mi = initial concentration (mol/L)Vi = initial volumeMf = final concentration (mol/L)Vf = final volume


Example: What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL?

Given: Vi = 25.0 mLMi = 6.00 MVf = 50.0 mL

Find: Mf

Vi x Mi = Vf x Mf


Vi x Mi = Vf x Mf

25.0 mL x 6.00 M = 50.0 mL x Mf

Mf = 25.0 mL x 6.00 M = 3.00 M50.0 mL

Note: Vi and Vf do not have to be in liters, but they must be in the same units.


Example: What volume of 0.50 M NaNO3 would be needed to prepare 100.0 mL of 0.10 M NaNO3 by dilution?

Given: Mi = 0.50 MVf = 100.0 mLMf = 0.10 M

Find: Vi

Vi x Mi = Vf x Mf


Vi x Mi = Vf x Mf

Vi x 0.50 M = 100.0 mL x 0.10 M

Vi = 100.0 mL x 0.10 M = 2.0 x 101 mL0.50 M

Concentration of Solutions Solution Stoichiometry

Recall that reactions occur on a mole to mole basis.For pure reactants, we measure reactants

using mass

For reactants that are added to a reaction as aqueous solutions, we measure the reactants using volume of solution.

Molarity

B

Molar

mass

Molarity

A


gramsB

gramsA

molesA

molesB

VolSoln A

Vol Soln B

Molar

mass

Molar ratio


Example: If 25.0 mL of 2.5 M NaOH are needed to neutralize (i.e. react completely with) a solution of H3PO4, how many moles of H3PO4 were present in the solution?

Given: 25.0 mL 2.5 M NaOH

Find: moles of H3PO4

3NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3H2O(l)

Molarity

B

Molarity

A


molesA

molesB

VolSoln A

Vol Soln B

Molar ratio

Molarity

NaOH


molesNaOH

molesH3PO4

VolNaOH Soln

Molar ratio


Strategy: mL NaOH soln L NaOH soln

mol H3PO4 mol NaOH


Mol H3PO4 = 25.0 mL x 1 L1000 mL

x 2.50 mol NaOH x 1 mol H3PO4

1 L3 mol NaOH

= 0.0208 mol H3PO4


Solution stoichiometry can be used to determine the concentration of aqueous solutions used in reactions.

Concentration of an acid can be determined using a process called titration.reacting a known volume of the acid

with a known volume of a standard base solution (i.e. a base whose concentration is known)


Example: If 35.50 mL of 2.5 M NaOH are needed to neutralize 50.0 mL of an H3PO4 solution, what is the concentration (molarity) of the H3PO4 solution?

Given: 35.50 mL 2.5 M NaOH50.0 mL of H3PO4 sol’n

Find: molarity (mol/L) H3PO4

3NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3H2O(l)


Strategy: M = moles L

To find the concentration of H3PO4 soln, we need both # moles and volume of H3PO4.

Since volume is given, we can simply find

moles and plug into the equation for M.

Molarity

NaOH


molesNaOH

molesH3PO4

VolNaOH Soln

Molar ratio


Mol H3PO4 = 35.5 mL x 1 L1000 mL

x 2.50 mol NaOH x 1 mol H3PO4

1 L 3 mol NaOH

= 0.0296 mol H3PO4

We’re not done….we need molarity.


Molarity = moles L

= 0.0296 mol H3PO4 x 1000 mL 50.0 mL L

= 0.592 M H3PO4

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Concentrations of Solutions