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UNIT 5. Aqueous Reactions and Solution Stoichiometry Molarity. Concentration of an Aqueous Solution - Molarity. An important property of a solution is its concentration : the amount of solute dissolved in a given quantity of solvent.

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UNIT 5

Aqueous Reactions and Solution Stoichiometry

Molarity

concentration of an aqueous solution molarity
Concentration of an Aqueous Solution - Molarity

An important property of a solution is itsconcentration:the amount of solute dissolved in a given quantity of solvent.

There are many ways to express concentration. We will learn one: molarity.

Molarity (symbol M) = moles of solute

volume of solution in liters

Molarity can be treated as a conversion factor (like density) to go from volume of a solution to moles of solute and vice versa.

slide3

Two Ways Calculate Moles

You now know two ways to calculate the number of moles of a substance.

1. If you have a known mass of the substance, you divide by the molar mass to get moles.

mass of A x 1 mol A = mol A

molar mass of A

2. If you have a known volume of a solution of known molarity, you multiply the volume of the solution in liters by the molarity to get moles.

liters of solution of A x molarity of A = mol A

slide4

Calculations Involving Molarity

How to prepare 1.00 liter of a 1.00 M solution of copper(II) sulfate:

1.00 L x 1.00 mole CuSO4 = 1.00 mole CuSO4

1 L solution

1.00 mole CuSO4 x 159.61 g CuSO4 = 160. g CuSO4

1 mole CuSO4

Put 160. g CuSO4 in a 1L volumetric flask. Fill about ¼ full with deionized (DI) water and swirl to dissolve. Fill to the mark with DI water and up end the flask to mix.

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Calculations Involving Molarity

How to prepare 250.0 mL of a 1.00 M solution of copper(II) sulfate:

250.0 mL x 1 L x 1.00 mole CuSO4 = 0.250 mole CuSO4

1000 mL 1 L solution

0.250 mole CuSO4 x 159.611 g CuSO4 = 39.9 g CuSO4

1 mole CuSO4

Put 39.9 g CuSO4 in a 250 mL volumetric flask and follow previous procedure.

Both flasks contain 1.00M CuSO4.

slide7

Calculations Involving Molarity

What is the molarity of a solution made by dissolving 25.0 g of NaCl in enough water to make 125 mL?

molarity = moles NaCl = 25.0 g NaCl x 1 mole NaCl

vol soln in L 58.44 g NaCl

125 mL x __1L___

1000 mL

= 0.428 moles NaCl

0.125 L

= 3.42 M

slide8

Molarity of Electrolytes

Molarity is just a unit that describes concentration.It can describe the concentration of the entire compound or any part of that compound.

A 0.500 M solution of potassium carbonate has what concentration of potassium ions? Carbonate ions?

0.500 M K2CO3 = 0.500 mol K2CO3 x 2 mol K+ = 1.00 M K+

1 L soln 1 mol K2CO3

0.500 M K2CO3 = 0.500 mol K2CO3 x 1 mol CO32- = 0.500 M CO32-

1 L soln 1 mol K2CO3

slide9

Using Molarity to Convert Between Moles and Volume

  • We have learned how to use the stoichiometry of a reaction to predict the yield of a product in grams or moles.
  • We can now extend this knowledge to include reactions involving solutions as reactants.
slide10

Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO3?

The steps are the same as with the stoichiometry we learned in Chapter 3.

1. Write and balance the equation for the reaction:

HNO3(aq) + KOH (aq)  H2O (l) + KNO3 (aq)

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Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO3?

2. Write what you know and what you need to know:

15.0 mL 6.0M ? mL 1.50M

HNO3(aq) + KOH (aq)  H2O (l) + KNO3 (aq)

3. Convert “what you know” to moles.

15.0 mL x 1 L x 6.0 mol HNO3

1000 mL 1.00 L solution

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Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO3?

15.0 mL 6.0M ? mL 1.50M

HNO3(aq) + KOH (aq)  H2O (l) + KNO3 (aq)

4. Convert to “moles of what you need to know” using the stoichiometry of the reaction:

15.0 mL x 1 L x 6.0 mol HNO3 x 1 mol KOH

1000 mL 1.00 L solution1 mol HNO3

slide13

Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO3?

15.0 mL 6.0M ? mL 1.50M

HNO3(aq) + KOH (aq)  H2O (l) + KNO3 (aq)

5. Convert your answer from moles to the required units.

We have been asked for volume. We have moles.

Molarity lets us convert between the two:

15.0 mL x 1 L x 6.0 mol HNO3 x 1 mol KOH x L of solution

1000 mL 1.00 L solution1 mol HNO3 1.50 mol KOH

60. mL of 1.50M KOH are required.

= 0.060 L

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Using Molarity to Convert Between Moles and Volume

Suppose the question had been “How many grams KOH are required to neutralize 15.0 mL of 6.0 M HNO3?” Only step 2 and step 5 change!

15.0 mL 6.0M ? g KOH (step 2)

HNO3(aq) + KOH (aq)  H2O (l) + KNO3 (aq)

5. Convert your answer from moles to the required units.

15.0 mL x 1 L x 6.0 mol HNO3 x 1 mol KOH x 56.11 g KOH

1000 mL 1.00 L solution1 mol HNO3 1 mol KOH

= 5.1 g KOH

slide15

Dilutions

liters of solution of A x molarity of A = mol A

This equality is helpful in dealing with dilutions.Often we are called upon to make a solution of a known concentrationnot from a dry chemical, but from another solution.

How do we make 200.0 mL of 2.50M HNO3 from concentrated (12.0 M) HNO3?

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Dilutions

How do we make 200.0 mL of 2.50M HNO3 from concentrated (12.0 M) HNO3?

200.0 mL x 1L x 2.50 mol HNO3 = 0.500 mol HNO3

1000 mL 1L

We must determine how many mLs of 12.0M HNO3 contains this same number of moles.

0.500 mol HNO3 x _____1L_____ = 0.0417 L

12.0 mol HNO3

Answer: 41.7 mL

slide17

Dilutions

How do we make 200.0 mL of 2.50M HNO3 from concentrated (12.0 M) HNO3?

The answer is to take 41.7 mL of 12.0 M HNO3 and add enough water to make 200.0 mL of solution…actually, you would add the acid to the water!

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Dilutions – A Quicker Way

How do we make 200.0 mL of 2.50M HNO3 from concentrated (12.0 M) HNO3?

If we set this up algebraically, we get

2.50M x 0.2000L = 12.0M x ?L.

With the answer included, we have

2.50M x 0.2000L = 12.0M x 0.0417L.

slide19

Dilutions – A Quicker Way

The general form for dilutions is

M1V1 = M2V2

where V = volume (mL or L, just keep them the same)

and M = molarity (NOT MOLES!!!)

slide20

Dilutions – Examples

M1V1 = M2V2 The Dilution Equation

How many mLs of 0.750M HCl are needed to make 150.0 mL of 0.125M HCl? How much water should be used?

(0.125M)(150.0mL) = (0.750M) (x mL)

x = 25.0 mL

Water used = 150.0 – 25.0 = 125.0 mL

slide21

Dilutions – Examples

M1V1 = M2V2

What is the concentration of a solution made by diluting 250.0 mL of 18M sulfuric acid to 1500.0 mL?

(18M)(250.0 mL) = (x M)(1500.0 mL)

x = 3.0M

slide22

Dilutions – Mixing Two Solutions

Often reactions are carried out by mixing two solutions (one for each reactant) together. It then becomes necessary to calculate the concentration of the reacting species in the resulting solution.

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

(NH4)3PO4(aq) + Na3PO4(aq) NR (no reaction…why?)

This is just a dilution!

slide23

Dilutions – Mixing Two Solutions

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

molarity = ________moles ion____________

volume of the final solution in liters

volume of the final solution = 350.0 + 125.0 = 475.0 mL

slide24

Dilutions – Mixing Two Solutions

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

final volume = 475.0 mL = 0.4750 L molarity = moles ion

0.4750 L

Let’s use the dilution equation to find the concentration of ammonium phosphate (NH4)3PO4 in the resulting solution:

350.0 mL (0.820 M) = 475.0 mL M2

M2 = 0.6042 mol (NH4)3PO4 per liter of the final solution

slide25

Dilutions – Mixing Two Solutions

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

final volume = 475.0 mL = 0.4750 L

Let’s use the dilution equation to find the concentration of sodium phosphate Na3PO4 in the resulting solution:

125.0 mL (0.750 M) = 475.0 mL M2

M2 = 0.1974 mol Na3PO4 per liter of the final solution

slide26

Dilutions – Mixing Two Solutions

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

Find the concentration of NH4+ in the final solution:

0.6042 mol (NH4)3PO4 x 3 mol NH4+ = 1.81 mol NH4+

L of final soln 1 mol (NH4)3PO4 L

1.81 mol NH4+ = 1.81 M NH4+

L

slide27

Dilutions – Mixing Two Solutions

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

Find the concentration of Na+ in the final solution:

0.1974 mol Na3PO4 x 3 mol Na+ = 0.592 mol Na+

L of final soln 1 mol Na3PO4 L

0.592 mol Na+ = 0.592 M Na+

L

slide28

Dilutions – Mixing Two Solutions

Example: 350.0 mL of 0.820 M ammonium phosphate are mixed with 125.0 mL of 0.750 M sodium phosphate. What is the concentration of each ion in the resulting solution?

The phosphate comes from both solutions. To find its concentration in the final solution, add the concentrations of (NH4)3PO4 and Na3PO4 in the final solution (why?):

M of PO43-in final solution

= M of (NH4)3PO4 + M of Na3PO4

= 0.6042 + 0.1974

= 0.802 M PO43-in final solution

Both molarities are for the final solution.

slide29

Titration

  • Titration is a way to determine the concentration of a solutionby reacting it with a known amount of a chemical(the standard)and monitoring the point of stoichiometric equivalence with anindicator.
  • Titration can be performed on any solution that gives a product which can be monitored:
    • acid-base titrations
    • redox titrations
    • precipitation titrations
slide30

Titration

  • Titration is a way to determine the concentration of a solutionby reacting it with a known amount of a chemical(the standard)and monitoring the point of stoichiometric equivalence with anindicator.
  • You have already performed the calculations for an acid-base titration. This is just a stoichiometric problem with the concentration of one of the reactants known.
slide31

Titration – Example 1

KHP (potassium acid phthalate) is a primary standard with a molar mass of 204.2 g. NaOH is a secondary standard, the concentration of which is found by titration with a known amount of KHP, according to the following equation:

15.44 mL ? M 0.396 g

NaOH (aq) + KHP (s)  KNaP (aq) + H2O(l)

If 15.44 mL of NaOH solution are needed to reach aphenolphthaleinendpoint in the titration of 0.396 g of KHP, what is the molarity of the NaOH solution?

slide32

Titration – Example 1

15.44 mL ? M 0.396 g

NaOH (aq) + KHP (s)  KNaP (aq) + H2O(l)

molarity = moles NaOH = 0.396 g KHP x 1 mol KHP x 1 molNaOH

L of solution 204.2 g KHP 1 mol KHP

15.44 mL x 1L

1000 mL

= 1.93 x 10-3molNaOH/0.01544 L soln

= 0.126 M NaOH

slide33

Titration – Example 2

The NaOH solution we just titrated is a secondary standard and can be used to titrate acid samples such as vinegar, a solution of acetic acid in water. If 25.00 mL of the vinegar is neutralized by 26.63 mL of the NaOH solution, what is the molarity of the acetic acid in the vinegar?

26.63 mL 0.126 M 25.00 mL ? M

NaOH (aq) + HC2H3O2 (aq)  NaC2H3O2 (aq) + H2O(l)

slide34

Titration – Example 2

26.63 mL 0.126 M 25.00 mL ? M

NaOH (aq) + HC2H3O2 (aq)  NaC2H3O2 (aq) + H2O(l)

The key to most titration problems is to keep track of which solution you need at what time.

molarity of acetic acid = moles acetic acid

volume of acetic acid (vinegar) in L

We get moles of acetic acid from our NaOH data:

26.63 mL NaOH x 1 L x 0.126 molNaOH x 1 mol acetic acid

1000 mL 1L NaOHsoln 1 molNaOH

= 3.36 x 10-3mol acetic acid

slide35

Titration – Example 2

26.63 mL 0.126 M 25.00 mL ? M

NaOH (aq) + HC2H3O2 (aq)  NaC2H3O2 (aq) + H2O(l)

molarity of acetic acid = 3.36 x 10-3mol acetic acid

volume of acetic acid (vinegar) in L

= 3.36 x 10-3mol acetic acid

25.00 mL x 1L

1000 mL

= 3.36 x 10-3mol acetic acid

0.02500 L

= 0.134 M

slide36

Titration – Example 3: Polyprotic Acids

If 25.00 mL of a solution of malonic acid (H2C3H2O4) is neutralized by 26.63 mL of the NaOH solution, what is the molarity of the malonic acid?

Note the position of the acid H’s in malonic acid.

26.63 mL 0.126 M 25.00 mL ? M

2NaOH (aq) + H2C3H2O4(aq)  Na2C3H2O4(aq) + 2H2O(l)

26.63 mL NaOH x 1 L x 0.126 molNaOH x 1 molmalonicacid

1000 mL 1L NaOHsoln2 molNaOH

= 1.68 x 10-3molmalonicacid

molarity of acetic acid = 1.68 x 10-3mol = 0.0671 M

0.02500 L