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Concentration Measurement:

Molarity

Instructor Dr. SAMIRA AMIR COGS Alfaisal University

Molarity

Molarity

Let’s start our discussion of solution concentration

Measurement with

Molarity: the most common solution concentration

measurement is defined as follows:

M =

M = moles of solute

moles of solute

LITERS

LITERS of

of solution

solution

Which means Molarity = # of moles per liter of solution

MOLARITY

MOLARITY

Capital M

Molarity

Molarity

Let’s Consider the following Problem:

What is the molarity of a solution in which 50g of

CuSO

4

is dissolved in water to make a 2 liter solution.

Step 1. What’s the formula for MOLARITY?

M = moles of solute

liters of solution

44

Molarity

Molarity

Step 2. We need to determine how many moles we have

in 50g of CuSO

4

.

How do we do that? Remember the mole Hole?

• 1. Determine the molar mass of CuSO

4

(159.5g/mol)

• 2. Calculate the # of moles of CuSO

4

Since we’re going into the mole hole we need

Since we’re going into the mole hole we need

to DIVIDE # of grams by molar mass

to DIVIDE # of grams by molar mass

(50/159.5) = 0.31 moles

Molarity

Molarity

Step 3. Calculate the Molarity of the solution using

the formula for MOLARITY

M = moles of Solute

liters of solution

How many moles do we have?

How many liters do we have?

So M = 0.31 moles

2 liters

M = 0.155 mol/liter

0.31 mol

2 L

Making a 1M NaCl Solution

Making a 1M NaCl Solution

1. Weigh out and add 1 mole of NaCl to a volumetric Flask
2. Add a small amount of solvent to dissolve the salt
   3. Once the solute
3. Once the solute

is dissolved add is dissolved add

additional

additional solvent

solvent

up to the

up to the 1 liter

1 liter

mark

mark

conc dil

Molarity by dilution

Molarity by dilution

Now what if we had a solution already prepared but

Wanted to decrease the molarity to some lower precise

Value.

How could we use an already prepared higher molarity

solution to accomplish this?

Note that it is only possible to go from a more

conc. (higher molarity) solution to a less conc.

(lower molarity) solution.

Molarity by dilution

Molarity by dilution

Recall that Molarity (M) = # moles of solute

liters of solution

So we need to dilute the solution by adding solvent

But how much more solvent do we need to add?

To make something more dilute we’re going to need to

add more solvent to the solution.

Molarity by dilution

Molarity by dilution

When we add more solvent to a solution do we

change the number of moles of solute in the

solution?

NO! But what does change?

We change the # of moles

per unit volume of Solvent because

the same # of moles are present but

now there’s more liquid

Molarity by Dilution

Molarity by Dilution

Let’s call the Molarity of the original solution M1 and the

New Molarity M

The number of liters of the original solution and the new

Solution will then be designated by V1 and V

So the molarity of the old solution is…

M1 = #moles of solute = #moles

#liters of solution V

So… #moles = M1 x V

Let’s see how this works …

Molarity by Dilution

Molarity by Dilution

The Molarity of the new solution will then be …

M2 = #moles of solute

#liters of solution

moles = M2 x V

Remember that the # of moles does not change when

We dilute solutions.

Molarity by dilution

Molarity by dilution

So…

M1 x V1 = M2 x V

If we know three values we can calculate the 4

th

.

Let’s see how we use this relationship

Dilution Problem

Dilution Problem

How much of a 3M solution of NiCL

2

is needed to

Prepare 100 ml of a 1.75M solution of NiCl

2

?

OK how do we solve this? What’s the equation

we’re going to use?

Right! M1V1 = M2V

M1 = 3M; M2 = 1.75M; V2 = 0.1L (100ml)

So … 3 x V1 = 1.75 x 0.

V1 = 0.058L or 58ml

Molarity by dilution

Molarity by dilution

How do we make 5L of

A 1.5M sol’n of KCL

From 12.0M stock sol’n?

M1V1 = M2V

12 x V1 = 1.5 x 5

12V1= 7.

V1 = 7.5/12 = 0.625L

Add 4.375L H

2

0 to a

0.625L of a 12M sol’n

Single Displacement

Single Displacement

and

and

Double Displacement

Double Displacement

Reactions

Reactions

1616

In

In single displacement

single displacement

reactions

reactions

, one element takes

, one element takes

the place of (displaces)

the place of (displaces)

another element in a

another element in a

compound.

compound.

general equations

general equations :

:

A + BX

A + BX 



AX + B

AX + B

or

or

AX + Y

AX + Y





AY + X

AY + X

example

example :

:

Mg

Mg

(s)

(s)

• 2HCl

• 2HCl

(aq)

(aq)





MgCl

MgCl

2(aq)

2(aq)

• H

• H

2(g)

2(g)

In a

In a

double displacement reaction

double displacement reaction

,

,

the cations of two different compounds

the cations of two different compounds

exchange places, forming two new

exchange places, forming two new

compounds.

compounds.

general equation

general equation :

:

WX + YZ

WX + YZ





WZ + YX

WZ + YX

example

example :

:

Ba(OH)

Ba(OH)

2(aq)2(aq)

• Na

• Na

22

SO

SO

4(aq)4(aq)





BaSO

BaSO

4(s)4(s)

+

+

2NaOH

2NaOH

(aq)(aq)

Acids and bases are chemical compounds.

Acids and bases are chemical compounds.

When they react together, they both lose

When they react together, they both lose

their properties. The reaction between an

their properties. The reaction between an

acid and a base is a special kind of double

acid and a base is a special kind of double

displacement reaction called

displacement reaction called neutralization

neutralization .

.

example:

example:

NaOH

NaOH

(aq)

(aq)

• HCl

• HCl

(aq)

(aq)





NaCl

NaCl

(aq)

(aq)

+ H

+ H

2

2

O

O

(

( l

l )

)

Practice Questions:

Practice Questions:

Write skeleton equations for the

Write skeleton equations for the

following. Then tell if the reaction is

following. Then tell if the reaction is

single or double displacement.

single or double displacement.

Finally, balance the equation.

Finally, balance the equation.

potassium + water

potassium + water





potassium

potassium

hydroxide + hydrogen gas

hydroxide + hydrogen gas

Skeleton equation

Skeleton equation

:

:

K + H

K + H

2

2

O

O





KOH + H

KOH + H

2

2

single or double displacement?

single or double displacement?

Single Displacement Reaction

Single Displacement Reaction

balanced equation:

balanced equation:

__K + __H

__K + __H

2

2

O

O





__KOH + __H

__KOH + __H

2

2

K +

K +

H

H

22

O

O





KOH + H

KOH + H

22

zinc + tin(II)chloride

zinc + tin(II)chloride





zinc

zinc

chloride + tin

chloride + tin

Skeleton equation:

Skeleton equation:

Zn + SnCl

Zn + SnCl





ZnCl+ Sn

ZnCl+ Sn

single or double displacement?

single or double displacement?

Single Displacement Reaction

Single Displacement Reaction

__Zn + __SnCl

__Zn + __SnCl 



__ZnCl + __Sn

__ZnCl + __Sn

Zn + SnCl

Zn + SnCl





ZnCl + Sn

ZnCl + Sn

(balanced)

(balanced)

The limiting reagent

(or limiting reactant) in a chemical reaction is the

substance that is totally consumed when the chemical

reaction is complete. The amount of product formed

is limited by this reagent,

since the reaction cannot continue without it.

If one or more other reagents are present in excess of the

quantities required to react with the limiting reagent, they

are described as excess reagents or excess reactants.

The limiting reagent must be identified in order to

calculate the percentage yield of a reaction, since the

theoretical yield is defined as the amount of product

obtained when the limiting reagent reacts completely.