Concentrations of Solutions

•Behavior of solutions depend on compound itself and on how much is present, i.e. on the concentration.•Two solutions can contain the same compounds but behave quite different because the proportions of those compounds are different.

Concentrations of Solutions

• Concentration of a solution: the more solute in a given volume of solvent, the more concentrated

• 1 tsp salt (NaCl)/cup of water

vs

• 3 Tbsp salt/cup water

Molarity

Molarity is one way to measure the concentration of a solution.

A 1.00 molar (1.00 M) solution contains 1.00 mol solute in every 1 liter of solution.

Units of molarity are: mol/L = M

moles of solute

volume of solution in litersMolarity (M) =

Preparing a 1.0 Molar Solution

One liter of a 1.00 M NaCl solution • need 1.00 mol of NaCl• weigh out 58.5 g NaCl (1.00 mole) and• add water to make 1.00 liter (total volume) of

solution.

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Molarity Practice

• What is the molar NaCl concentration if you have 0.5 mol NaCl in 1.00 L of solution?

• What is the molar NaCl concentration if you have 0.5 mol NaCl in 0.50 L of solution?

0.5 mol NaCl/1.00 L = 0.5 mol/L = 0.5 M

0.5 mol NaCl/0.50 L = 0.5/0.50 mol/L = 1 mol/L = 1 M

Molarity Practice

What is the molar NaCl concentration if you have 10.0 g of NaCl in 1.00 L of solution?

Have grams not mols!• Grams mol • Need molar mass

NaCl: 23.00 + 35.45= 58.45 g/molSo: 10.0g x (1 mol/35.45g)=0.282 mol NaClMolar concentration: 0.282 mol/1.00 L = 0.282 M

Molarity – Moles - Volume

• Have mol and vol molarity

• Have molarity and vol mol of solute

• Have molarity and mol of solute volume

• AND: mol of solute grams of solute

moles of solute

volume of solution in litersMolarity (M) =

mol

Volume (L)Molarity (M) =

Practice

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

Given: 2.5 L of soln

0.10M HCl

Find: mol HCl

Use: mol = molarity x volume

mol HCl =0.10 M HCl x 2.5 L 0.10 mol HCl=1 L

x 2.5 L

= 0.25 mol HCl

= 0.10 mol/1 L HCl

PracticeWhat 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 NaOH

Find: vol soln

Use: vol soln = mol solute / molarity

= 0.10 mol NaOH / 1L

Vol soln = 0.50 mol NaOH0.10 M NaOH

= 0.50 mol NaOH0.10 mol NaOH

1L

= 0.50 mol NaOH X 1L0.10 mol NaOH = 5 L

More PracticeHow many grams of CuSO4 are needed to prepare 250.0 mL of 1.00 M CuSO4?

Given: 250.0 mL soln

1.00 M CuSO4

Find: g CuSO4

Use: mol CuSO4 = molarity x volume

Molarity = mol / 1L

Vol = 250.0 mL

Concentration of Solutions Interconverting Molarity, Moles, and Volume

g CuSO4 = 250.0 mL soln x 1 L x 1.00 mol

1000 mL 1 L soln

x 159.6 g CuSO4

1 mol

= 39.9 g CuSO4

Steps involved in preparing solutions from pure solids

Steps involved in preparing solutions from pure solids

– Calculate the amount of solid required– Weigh out the solid– Place in an appropriate volumetric flask– Fill flask about half full with water and mix.– Fill to the mark with water and invert to mix.

You should be able to describe this process (including calculating the mass of solid to use) for any solution I specify.

Dilutions

• Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions).e.g. 12 M HCl

12 M H2SO4

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

Dilutions

• A given volume of a stock solution contains a specific number of moles of solute.

e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl

(How do you know this???)

• 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

0.15 mol

25 mL HCl

+

25 mL H2O

=50 mL0.15 mol

Dilutionsmoles solute = moles solute

before 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 = moles solute

Volume of solution

Dilution Calculation

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

Mc x Vc = Md x Vd

where Mc= initial concentration (mol/L)= more

concentrated

Vc = initial volume of more conc. solution

Md =final concentration (mol/L) in dilution

Vd = final volume of diluted solution

Dilution Calculation

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: Vc = 25.0 mL

Mc = 6.00 M

Vd = 50.0 mL

Find: Md

Use Vcx Mc= Vdx Md Solve for Md

Note: Vcand Vd do not have to be in liters, but they must be in the same units.

Dilution• Make a diluted solution once you know Vc and Vd

– Use a pipet to deliver a volume of the concentrated solution to a new volumetric flask.

– Add solvent to the line on the neck of the new flask. – Mix well.

Practice

• How many mL of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution?

• If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting solution?

Vc = ? Mc = 5.0M Md = 0.10MVd = 250 mL

Vc = 10.0 mLMd = ? Mc = 10.0M Vd = 250 mL

Solution Stoichiometry

• Remember: 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

Solution Stoichiometry

gramsB

gramsA

molesA

molesB

VolSoln A

Vol Soln B

Molar

mass

Molar ratio

Solution Stoichiometry Practice

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

balanced eqn: 3 mol NaOH/1 mol H3PO4

Find: moles of H3PO4

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

Molarity

NaOH

Approach

molesNaOH

molesH3PO4

VolNaOH Soln

Molar ratio

25.0 mL NaOH soln

0.025 L NaOH soln

2.5 M (=mol/L)

3 mol NaOH/1 mol H3PO4

Mol NaOH = 25.0 mL x 1000 mL

1L

1 L

2.5 molx = 0.0625 mol NaOH

More practice

What mass of aluminum hydroxide is needed to neutralize 12.5 mL of 0.50 M sulfuric acid?

Solution Stoichiometry

• 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)

Titration

Practice

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 NaOH

50.0 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

Plan

molesNaOH

molesH3PO4

VolNaOH Soln

Molar ratio

Mol H3PO4 = 35.5 mL x 1 L

1000 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 of H3PO4

Molarity = moles

L

= 0.0296 mol H3PO4 x 1000 mL

50.0 mL L

= 0.592 M H3PO4