section 6.2—concentration how do we indicate how much of the electrolytes are in the drink?...

Section 6.2—Concentration

How do we indicate how much of the electrolytes are in the drink?

Objectives:•Calculate concentrations in percent concentration and molarity•Calculate the concentration of electrolytes in solutions of ionic compounds

Concentrated versus Dilute

solute solvent

Lower concentration

Not as many solute particles

Higher concentrationMore solute particles

Concentration

Concentration gives the ratio of amount of solute dissolved to total amount of solute and solvent

There are several ways to show concentration

Percent Mass/Volume

This is a method of showing concentration that is not used as often in chemistry

However, it’s used often in the food and drink industryFor example, your diet drink can might say you

have less than 0.035 g of salt in 240 mL. That would give you a concentration of

0.035 g / 240 mL, which is 0.015% solution

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%(M /V ) =grams solute

mL solvent×100

%(M/V) Example

Example:If you dissolve 12 g of sugar in 150 mL

of water, what percent

mass/volume is the solution?

€

%(M /V ) =grams solute

mL solvent×100

%(M/V) Example

€

%(M /V ) =12g

150mL×100

Example:If you dissolve 12 g of sugar in 150 mL

of water, what percent

weight/volume is the solution?

€

%(M /V ) =grams solute

L solvent×100

8.0% (M/V)

Practice Problems

Determine the % M/V of a carbohydrate in a sports drink if it contains 14 g in 240 mL; if it contains 20 g in 240 mL.

A certain sports drink contains 0.050 g sodium chloride in 240 mL. What is its concentration in % M/V? What if it had 0.100 g in 240 mL?

Concentration using # of molecules

When working with chemistry and molecules, it’s more convenient to have a concentration that represents the number of molecules of solute rather than the mass (since they all have different masses)

Remember, we use moles as a way of counting molecules in large numbers

Mole Review

The molar mass of a substance is found by adding up all the atomic masses in the substanceExample: The molar mass of NaCl is 58.5 g

The molar mass of a substance in grams represents 1 mole of that substance. 1 mole of NaCl has a mass of 58.5 g

Practice

Determine the molar mass of each of the following substances.Ca(OH)2

Mg

C2H3OH

Cu(NO3)2

Review: Calculating # of Moles

If you are given the mass of a substance, you use the molar mass to determine the number of moles.Example how many moles are in 35 g of Mg?35 g x 1 mole = 1.4 moles

24.3 g

Practice:

How many moles are in 25.5 g NaCl?

25.5 g NaCl = _______ mole NaClg NaCl

mole NaCl1

58.5

0.436

1 mole NaCl molecules = 58.5 g

NaCl

11

23.0 g/mole35.5 g/mole

= 23.0 g/mole= 35.5 g/mole+

58.5 g/mole

Example:How many moles are in 25.5 g NaCl?

More Practice

Determine the number of moles in each of the following:

1.68.0 g CaCl2

2.9.2 g CH4

3. 70.1 g Al(OH)3

Molarity

Molarity (M) is a concentration unit that uses moles of the solute instead of the mass of the solute

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M =moles solute

L solution

Molarity ExampleExample:

If you dissolve 12 g of NaCl to make

150 mL of solution, what is the molarity?

Molarity Example

L

molesM

150.0

21.0

Example:If you dissolve 12 g

of NaCl to make 150 mL of solution,

what is the molarity?

€

M =moles solute

L solution= 1.4 M NaCl

12 g NaCl = _______ mole NaClg NaCl

mole NaCl1

58.5

0.21

1 mole NaCl molecules = 58.5 g

NaCl

11

23.0 g/mole35.5 g/mole

= 23.0 g/mole= 35.5 g/mole+

58.5 g/mole

Next, remember to change mL to L! Divide by 1000.150 mL of water = 0.150 L

Practice Problems

1. Determine the molarity of 2.5 L of solution made from 125 g of AgNO3.

2. What is the molarity of a solution produced when 145 g of NaCl is dissolved in enough water to make 2.75 L of solution?

3. What is the molarity of a solution in which 85.6 g of HCl are dissolved in enough water to make 385 mL of solution?

4. 8.77 g of KCl is dissolved in sufficient water to make 4.75 L of solution. Determine the molarity of the solution.

Converting between the two

If you know the %(W/V), you know the mass of the solute

You can convert that mass into moles using molecular mass

You can then use the moles solute to find molarity

Converting from % to M Example

Example:What molarity is a 250 mL sample of 7.0 %(W/V) NaCl?

Converting from % to M Example

L

molesM

250.0

30.0

Example:What molarity is a 250 mL sample of 7.0 %(W/V) NaCl?

1.2 M NaCl

17.5 g NaCl = _______ mole NaClg NaCl

mole NaCl1

58.44

0.30

1 mole NaCl molecules = 58.44 g

NaCl

11

22.99 g/mole35.45 g/mole

= 22.99 g/mole= 35.45 g/mole+

58.44 g/mole

Remember to change mL to L! 250 mL of water = 0.250 L

100250

?)/%(0.7

mL

gVW ? = 17.5 g NaCl

Concentration of Electrolytes

An electrolyte breaks up into ions when dissolved in water

You have to take into account how the compound breaks up to determine the concentration of the ions

CaCl2 Ca+2 + 2 Cl-1

For every 1 CaCl2 unit that dissolves, you will produce 1 Ca+2 ion and 2 Cl-1 ions

If the concentration of CaCl2 is 0.25 M, the concentration of Ca+2 is 0.25 M and Cl-1 is 0.50 M

Let’s Practice #1

Example:You want to make 200 mL of a 15% (W/V) solution of

sugar. What mass of sugar do you

need to add to the water?

100)/%( solventmL

solutegramsVW

Let’s Practice #1

100200

?)/%(15

mL

gVW

Example:You want to make 200 mL of a 15% (W/V) solution of

sugar. What mass of sugar do you

need to add to the water?

100)/%( solventmL

solutegramsVW

30 g of sugar

Let’s Practice #2

Example:What is the %(W/V)

of a 500. mL sample of a 0.25 M

CaCl2 solution?

Let’s Practice #2

L

molesM

500.0

?25.0

Example:What is the %(W/V)

of a 500. mL sample of a 0.25 M

CaCl2 solution?

2.8 %(W/V) CaCl2

0.125 moles CaCl2 = _______ g CaCl2

mole CaCl2

g CaCl2110.98

1

13.9

1 mole CaCl2 molecules = 110.98 g

CaCl

12

40.08 g/mole35.45 g/mole

= 40.08 g/mole= 70.90 g/mole+

110.98 g/mole

100500

9.13)/%(

mL

gVW

? = 0.125 moles CaCl2

Let’s Practice #3

Example:What are the

molarities of the ions made in a 0.75 M

solution of Ca(NO3)2

Let’s Practice #3

Example:What are the

molarities of the ions made in a 0.75 M

solution of Ca(NO3)2

Ca(NO3)2 Ca+2 + 2 NO3-1

For every 1 Ca(NO3)2, there will be 1 Ca+2 and 2 NO3-1 ions

Ca+2 = 0.75 MNO3

-1 = 1.5 M