the chemistry of acids and bases. there she blows!!!

There she blows!!!There she blows!!!

Acids and Bases: Acids and Bases: Molar Molar SolutionsSolutionsAt the conclusion of our time At the conclusion of our time together, you should be able to:together, you should be able to:

1. Define molarity2. Make a solution with a given molarity3. Determine the molarity of a given solution

Some DefinitionsSome Definitions

A solution is a A solution is a homogeneous homogeneous mixture of 2 or mixture of 2 or more substances more substances in a single phase. in a single phase.

The larger The larger constituent is constituent is usually regarded usually regarded as the as the SOLVENTSOLVENT and the others as and the others as SOLUTESSOLUTES..


The units are The units are moles moles per Liter.per Liter.

Specifically, it’s Specifically, it’s moles of solute moles of solute per Liter of per Liter of solution.solution.

Abbreviated as Abbreviated as mol/Lmol/L or or MM..

The word The word MolarityMolarity..


Moles Moles measuremeasure amount of solute amount of solute usually in gramsusually in grams

Molarity Molarity measuresmeasures moles of solute moles of solute per Liter of per Liter of solution.solution.

What Do You Get From a What Do You Get From a Pampered Cow?Pampered Cow?

Spoiled Milk.

Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a solution is The amount of solute in a solution is given by its given by its concentrationconcentration.

Molarity(M)= moles soluteliters of solution


Therefore: x both sides by Liters =

MV = moles = grams/molar mass


PROBLEM: Using grams/molar mass = PROBLEM: Using grams/molar mass = MV.MV.

Handout #1Handout #1



= 58.44 g NaCl

x g NaCl

58.44 g NaCl= 1 M x 1 L





= 42.5 g NaNO3

x g NaNO3

85.00 g NaNO3

= 0.5 M x 1 L

Over-Worked MouseOver-Worked Mouse

Acids and Bases: Acids and Bases: Molar Molar SolutionsSolutionsLet’s see if you can:Let’s see if you can:

1. Define molarity2. Make a solution with a given molarity


Backside #1Backside #1



= 245 g H2SO4

x g H2SO4

98.09 g H2SO4

= 2.50 M x 1 L

Hopefully, you’re not this lost!!!

Interesting Vanity Plate…Interesting Vanity Plate…





= 0.741 g Ca(OH)2

x g Ca(OH)2

74.10 g Ca(OH)2

= 0.100 M x 0.100 L

Preparing SolutionsPreparing SolutionsPreparing SolutionsPreparing Solutions

Determine the mass of Determine the mass of solute.solute.

Place in the appropriate Place in the appropriate volumetric flask.volumetric flask.

Add deionized water and Add deionized water and swirl until solute is dissolved.swirl until solute is dissolved.

Add water to the mark on Add water to the mark on the neck of the flask.the neck of the flask.

Stopper and mix thoroughly.Stopper and mix thoroughly.

Support Bacteria!!Support Bacteria!!

They're the only culture some people have.

Stoichiometry: Molar SolutionsStoichiometry: Molar SolutionsLet’s see if you can:Let’s see if you can:



Therefore:




Therefore:


Now let’s use this formula to solve another type of problem:


Kid’s Letters Kid’s Letters to God:to God:

PROBLEM: PROBLEM: Mr. T accidentally dropped Mr. T accidentally dropped 15.14 g of silver (I) nitrate into 100.0 mL 15.14 g of silver (I) nitrate into 100.0 mL of deionized water. Rather than throw of deionized water. Rather than throw

the solution away, help him by the solution away, help him by determining its molar concentration so determining its molar concentration so

that he can label it and still use it.that he can label it and still use it...

PROBLEM: PROBLEM: Mr. T accidentally dropped Mr. T accidentally dropped 15.14 g of silver (I) nitrate into 100.0 mL 15.14 g of silver (I) nitrate into 100.0 mL of deionized water. Rather than throw of deionized water. Rather than throw

the solution away, help him by the solution away, help him by determining its molar concentration so determining its molar concentration so

that he can label it and still use it.that he can label it and still use it...

= 0.8912 M AgNO3

15.14 g AgNO3 x 1 mol AgNO3

169.88 g AgNO3

= M x 0.100 L

PROBLEM: PROBLEM: Molarity Problems #2Molarity Problems #2PROBLEM: PROBLEM: Molarity Problems #2Molarity Problems #2

= 3 M KI

249 g KI x 1 mol KI

166.00 g KI= M x 0.5 L

How Would You Make this How Would You Make this 3M Solution??3M Solution??

How Would You Make this How Would You Make this 3M Solution??3M Solution??

Measure 249 g of KIMeasure 249 g of KIPlace in a 0.500 L Place in a 0.500 L

volumetric flask.volumetric flask.Add deionized water and Add deionized water and

swirl until solute is swirl until solute is dissolved.dissolved.



Most Caring Person!!Most Caring Person!!

An elderly gentleman had recently lost his wife.Upon seeing the man cry, a little 4 year old neighbor boy went into the old gentleman's yard, climbed onto his lap, and just sat there.

When his mother asked him what he had said to the neighbor, the little boy just said,

"Nothing, I just helped him cry."

Acids and Bases: Acids and Bases: Molar Molar SolutionsSolutionsLet’s see if you can:Let’s see if you can:


How We How We Doing?? Need Doing?? Need Help????Help????

Exit Quiz #1Exit Quiz #1

1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water.

2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution.

1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water.

2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution.

0.75 mol NaCl

= X M x 0.3000 L

= 2.5 M NaCl

X mol HCl

= 2.45 M x 0.0122 L

= 0.0299 mol HCl

0.0299 mol HCl x 36.46 g HCl

1 mol HCl= 1.09 g HCl

Entrance Quiz #2 Entrance Quiz #2

1. Calculate the molarity if 0.50

mol of KBr in placed in 750 mL of water.

2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl in solution.

1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water.

2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl.

0.50 mol KBr

= X M x 0.75 L

= 0.67 M KBr

0.0672 mol HCl

= X M x 0.0122 L

= 5.51 M HCl

2.45 g HCl x 1 mol HCl

36.46 g HCl= 0.0672 mol HCl

Stoichiometry: Molar SolutionsStoichiometry: Molar SolutionsAt the conclusion of our time At the conclusion of our time together, you should be able to:together, you should be able to:

1. Define molarity2. Determine the molarity of a given solution3. Make a solution with a given molarity4. Dilute a given solution to a new molarity

If the Amount (moles) of If the Amount (moles) of Solute #1 = #2Solute #1 = #2

If the Amount (moles) of If the Amount (moles) of Solute #1 = #2Solute #1 = #2

M1V1 = moles and M2V2 = moles

Therefore if moles of solute are constant:

M1V1 = M2V2


And Using the Formula:

The Scientific MethodThe Scientific Methodbegins with begins with

Questions about the World Around Questions about the World Around You.You.

Ever Wonder Why?...Ever Wonder Why?...

there are flotation devices under plane seats instead of parachutes?

PROBLEM #1: Using PROBLEM #1: Using MM11VV11 = M = M22VV22 PROBLEM #1: Using PROBLEM #1: Using MM11VV11 = M = M22VV22

= 0.125 M NaOH

0.150 M NaOH x 0.125 L

= M x 0.150 L

PROBLEM #5: Using PROBLEM #5: Using MM11VV11 = M = M22VV22 PROBLEM #5: Using PROBLEM #5: Using MM11VV11 = M = M22VV22

= 1.20 L

Therefore: 0.700 L needs to be added

2.40 M KCl x 0.500 L

= 1.00 M KCl x X L

Stoichiometry: Molar SolutionsStoichiometry: Molar SolutionsLet’s see if you can:Let’s see if you can:

1. Define molarity2. Determine the molarity of a given solution3. Make a solution with a given molarity4. Dilute a given solution to a new molarity

I knew I shouldn’t have done that!!I knew I shouldn’t have done that!!

How much water do I need to add to 250 How much water do I need to add to 250 mL of 3.0 M HCl to dilute it to 1.0 M HCl?mL of 3.0 M HCl to dilute it to 1.0 M HCl?How much water do I need to add to 250 How much water do I need to add to 250 mL of 3.0 M HCl to dilute it to 1.0 M HCl?mL of 3.0 M HCl to dilute it to 1.0 M HCl?

= 0.75 L Total, therefore 0.50 L

3.0 M HCl x 0.250 L

= 1.0 M HCl x L

Name ________Class Period _____ Clicker Number

Name ________Class Period _____ Clicker Number

"Making Molar Solutions A1"

(10 points)

Make 50.00 mL of a 0.100M BaCl22H2O solution.

__________ grams mass of solute needed

__________ Instructor initials (one point)

What are the products of the reaction? Balance the equation. Write the balanced molecular, complete ionic and net ionic equations below. Place the precipitate on the 3rd line of the first row of lines. (5 points)

________ + ________ ________ + ________ ___ ____ + ___ ____ ______ + ___ ____ ____ + ___ _____ What is this the best molar ratio based on

the stoichiometry? Should you have put all 50.00 mL of each reactant together to form the most product? Circle yes or no. (2 points)

____________ : ____________ Yes No

Preparing SolutionsPreparing SolutionsPreparing SolutionsPreparing Solutions

Determine the mass of Determine the mass of solute.solute.

Place in the appropriate Place in the appropriate volumetric flask.volumetric flask.

Add deionized water and Add deionized water and swirl until solute is dissolved.swirl until solute is dissolved.



PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O O in enough water to make 250 mL of in enough water to make 250 mL of

solution. Calculate the Molarity.solution. Calculate the Molarity.

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O O in enough water to make 250 mL of in enough water to make 250 mL of

solution. Calculate the Molarity.solution. Calculate the Molarity.

Step 1: Step 1: Calculate moles of Calculate moles of NiClNiCl22•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 H22O O ] = 0.0841 M

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?

X g H2C2O4 x 1 mol H2C2O4

90.04 g H2C2O4

= 0.0500 x 0.250 L

= 1.13 g H2C2O4

Learning CheckLearning Check

How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

moles = MVmol = 3.0 M x 0.400 Lmol = 1.2

48 g NaOH

1.2 mol NaOH x 40.00 g NaOH

1 mol NaOH

PROBLEM: PROBLEM: Mr. T needs to make 0.500 L Mr. T needs to make 0.500 L of a 0.100 M solution of lead (II) nitrate of a 0.100 M solution of lead (II) nitrate for a future lab. Please help him by for a future lab. Please help him by calculating the amount of solute needed calculating the amount of solute needed and by outlining in detail the steps he and by outlining in detail the steps he would need to take to make this molar would need to take to make this molar solution.solution.

PROBLEM: PROBLEM: Mr. T needs to make 0.500 L Mr. T needs to make 0.500 L of a 0.100 M solution of lead (II) nitrate of a 0.100 M solution of lead (II) nitrate for a future lab. Please help him by for a future lab. Please help him by calculating the amount of solute needed calculating the amount of solute needed and by outlining in detail the steps he and by outlining in detail the steps he would need to take to make this molar would need to take to make this molar solution.solution.

= 16.6 g Pb(NO3)2

X g Pb(NO3)2 x 1 mol Pb(NO3)2

331.22 g Pb(NO3)2

= 0.100 M x 0.500 L

PROBLEM # 4: Grams to make 100. mL PROBLEM # 4: Grams to make 100. mL of 0.100 M calcium hydroxideof 0.100 M calcium hydroxide..PROBLEM # 4: Grams to make 100. mL PROBLEM # 4: Grams to make 100. mL of 0.100 M calcium hydroxideof 0.100 M calcium hydroxide..

= 0.741 g Ca(OH)2

X g Ca(OH)2 x 1 mol Ca(OH)2

74.10 g Ca(OH)2

= 0.100 M x 0.100 L

PROBLEM # 5: 4.00 moles of nitric acid PROBLEM # 5: 4.00 moles of nitric acid in 1.50 L of solution = what Min 1.50 L of solution = what M..PROBLEM # 5: 4.00 moles of nitric acid PROBLEM # 5: 4.00 moles of nitric acid in 1.50 L of solution = what Min 1.50 L of solution = what M..

= 2.67 M HNO3

4.00 mol HNO3

= X M x 1.50 L

the chemistry of acids and bases. there she blows!!!

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