Solution Concentration

Review

A solution is a homogeneous mixture.The solvent is the major component of the

solution.The solute is the minor component and

active ingredient.A saturated solution holds the maximum

amount of solute that is theoretically possible for a given temperature.

How would you describe this picture?

Solution Concentration

Is one glass of tea stronger than the other?– What’s true about the “stronger” glass of tea?– How much tea does it have in it compared to

the other glass?

Solution Concentration

Concentration – a ratio comparing the amount of solute to the amount of solution.

Many ways of expressing concentration:– % by weight (% w/w)– % by volume (% v/v)– parts per million (ppm) or parts per billion

(ppb) for very low concentrations– molality (m)– Molarity (M)

Concentrated vs. Dilute

The words “concentrated” and “dilute” are opposites.

EX: The dark tea is more concentrated than the light tea.

EX: The light tea is more dilute than the dark tea.

Concentrated vs. Dilute

Concentrated solution

Dilute solution

= solute particles

Percent by Weight

% by weight (% w/w)

What is the % w/w of a solution if 3.00 grams of NaCl are dissolved in 17.00 g of water?– mass of solute = 3.00 g

– mass of solution = 3.00 g + 17.00 g = 20.00 g

– (3.00 g / 20.00 g) x 100% = 15.0% w/w

100% x solution of mass total

solute of mass %w/w

Percent by Volume

% by volume (% v/v)

What is the % v/v of a solution if 20.0 mL of alcohol are dissolved in 50.0 mL of solution?– volume of solute = 20.0 mL

– volume of solution = 50.0 mL

– (20.0 mL / 50.0 mL) x 100% = 40.0%

100% x solution of volume total

solute of volume %v/v

Molarity

Molarity (M)

– UNITS: mol/L or Molar (M)

– Example: 0.500 mol/L = 0.500 M

solution of Liters

solute of moles Molarity

Molarity

What is the Molar concentration of a sol’n if 20.0 grams of KNO3 (MM = 101.11 g/mol) is dissolved in enough water to make 800. mL?

– Convert g of KNO3 to mol of KNO3

– Convert mL to L

33

33 KNO mol 0.198

KNO g 101.11

KNO mol 1 x KNO g 20.0

L 0.800 mL 1

L 0.001 x mL 800.

Molarity

What is the Molar concentration of a sol’n if 0.198 mol KNO3 is dissolved in enough water to make 0.800 L?

M 0.248L 0.800

KNO mol 0.198 Molarity 3

Molarity

What is the Molar concentration of a sol’n if 10.5 grams of glucose (MM = 180.18 g/mol) is dissolved in enough water to make 20.0 mL of sol’n?– Convert g of glucose to mol of glucose.

– Convert mL to L.

glucose mol 0.0583 glucose g 180.18

glucose mol 1 x glucose g 10.5

L 0.0200 mL 1

L 0.001 x mL 20.0

Molarity

What is the Molar concentration of a sol’n if 0.0583 mol of glucose is dissolved in enough water to make 0.0200 L of sol’n?

M 2.92 L 0.0200

glucose mol 0.0583 Molarity

Calculating Grams

How many grams of KI (MM = 166.00 g/mol) are needed to prepare 25.0 mL of a 0.750 M solution?

– Convert mL to L.

– Solve for moles.

– moles of KI = 0.750 M x 0.0250 L = 0.0188 mol KI

L 0.0250 mL 1

L 0.001 x mL 25.0

L 0.0250

KI of moles x M 0.750

Calculating Grams

How many grams of KI (MM = 166.00 g/mol) are needed to prepare 25.0 mL of a 0.750 M solution?

– Convert 0.0188 mol KI to grams.

KI g 3.12 KI mol 1

KI g 166.00 x KI mol 0.0188

Calculating Grams

How many grams of HNO3 (MM = 63.02 g/mol) are present in 50.0 mL of a 1.50 M sol’n?– Convert mL to L.

• 50.0 mL = 0.0500 L

– Solve for moles:• moles = (1.50 M)(0.0500 L) = 0.0750 mol HNO3

– Convert 0.0750 mol HNO3 to grams:

• 0.0750 mol HNO3 = 4.73 g HNO3

Dilution

Dilute (verb) - to add solvent to a solution.– Decreases sol'n concentration.

– M1V

1 = M

2V

2

• M1 = initial conc.

• V1 = initial volume

• M2 = final conc.

• V2 = final volume

– Assumes no solute is added.

Dilution

Stock SolutionImpractically

High Concentration

Usable Solution

Add H2O

Question for Consideration: Why do you think chemical supply companies typically sell acids (and other solutions) in extremely high concentrations when it would be safer to ship more dilute solutions?

Dilution

To what volume should 40.0 mL of 18 M H2SO

4

be diluted if a concentration of 3.0 M is desired?– What do we want to know?

• V2

– What do we already know?• M

1 = 18 M

• V1 = 40.0 mL

• M2 = 3.0 M

– (18 M)(40.0 mL) = (3.0 M)V2

– 720 M*mL = (3.0 M)V2

– V2 = 240 mL

Dilution

You are asked to prepare 500. mL of 0.250 M HCl, starting with a 12.0 Molar stock sol'n. How much stock should you use?– What do we want to know?

• V1

– What do we already know?• M

1 = 12.0 M

• M2 = 0.250 M

• V2 = 500. mL

– (12.0 M) V1 = (0.250 M)(500. mL)

– (12.0 M) V1 = 125 M*mL

– V1 = 10.4 mL

To how much water should you add 20.0 mL of 5.00 M HNO

3 to dilute it to 1.00 M?

– What do we want to know?• How much water to add. (V

2 - V

1)

– What do we already know?• M

1 = 5.00 M

• V1 = 20.0 mL

• M2 = 1.00 M

– (5.00 M)(20.0 mL) = (1.00 M) V2

– 100. M*mL = (1.00 M) V2

– V2 = 100. mL

– Water added = 100. mL - 20.0 mL = 80. mL

Dilution

Colligative Properties

Colligative properties are properties of solutions that are affected by the number of particles but not the identity of the solute

Boiling Point Elevation: A Colligative Property

Boiling point elevation is the temperature difference between a solution and pure solvent

The value of the boiling point elevation is directly proportional to molality, meaning the greater the number of solute particles, the greater the elevation

Freezing Point Depression: A Colligative Property

Freezing point depression is the difference in temperature between a solution and a pure solvent

The value of freezing point depression is directly proportional to molality, meaning the greater the number of solute particles