Solution A homogeneous mixture of two or more substances.

Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. For Any Solution

Solvent Solvent - there can be only ONE Solute Solute - there can be MORE than one For Any Solution

Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What is the solvent in air? in air?

Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What are the solutes in air? in air?

Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What is the solvent in stainless steel? in stainless steel?

Solvent Solvent - the substance present in the most amount. Solute Solute - the substance present in the least amount. What are the solutes in stainless steel? in stainless steel?

Solvent Solvent - Water Solute Solute - The substance dissolved in water Aqueous Solution

The amount of solute that can be dissolved in a given amount of solvent, at a given temperature. Solubility

Is NaCl soluble in H 2 O?

A solution containing the maximum amount of a solute that is possible to dissolve in a given volume of solvent at a given temperature. Saturated Solution

NaCl has a solubility of 357 grams per liter of "cold" H 2 O. Saturated Solution +

How many moles of NaCl are in the solution? Saturated Solution NaClsolution

357 g NaCl 1 mole NaCl Saturated Solution NaClsolution 58 g NaCl 6.2 mole NaCl =

How many "particles" of NaCl are in the solution? Saturated Solution NaClsolution

357 g NaCl Saturated Solution NaClsolution 58 g NaCl 3.7 X particles NaCl = 6.02 X particles NaCl

A comparison of the amounts of solute and solvent in a solution. Concentration

"Strong" and "Weak" give SOME comparison, but only a general idea. Concentration

"Dilute" and "Concentrated" still don't provide enough for quantitative calculations. Concentration

To do calculations, we must know "how much" solute and "how much" solvent are present. Concentration

M = Molarity moles solute dm 3 solution

1 mole = formula mass (g) 1 liter = 1 cubic decimeter dm 3 1 liter = 1000 milliliters 1 ml = 1 cm 3

1. What is the molarity of a liter of solution containing 100 grams of copper (II) chloride? Molarity = moles dm 3

1 0 0 g C u C l 2 1 l i t e r Molarity = moles dm 3

1 0 0 g C u C l 2 1 l i t e r Molarity = grams moles dm 3

1 0 0 g C u C l 2 1 l i t e r Molarity = grams mole moles dm 3

1 0 0 g C u C l 2 1 l i t e r Molarity = grams 1 mole moles dm 3

1 0 0 g C u C l 2 1 m o l e 1 l i t e r g Molarity = Cu = 1 X 64 = 64 Cl = 2 X 35 = moles dm 3

1 0 0 g C u C l 2 1 m o l e 1 l i t e r 1 l i t e r g 1 d m 3 Molarity = moles dm 3

1 0 0 g C u C l 2 1 m o l e 1 l i t e r 1 l i t e r g 1 d m 3 Molarity = moles dm 3 Have we worked the problem?

1 0 0 g C u C l 2 1 m o l e 1 l i t e r 1 l i t e r g 1 d m M C u C l 2 Molarity = moles dm 3

2.How much NaCl is needed to prepare 250ml of 0.5M salt water How is this problem different from the first?

1. What is the molarity of a liter of solution containing 100 grams of copper (II) chloride? 2.How much NaCl is needed to prepare 250ml of 0.5M salt water? The second GIVES M The first ASKS for M

2.How much NaCl is needed to prepare 250 ml of 0.5 M salt water

250 ml 0.5 mole NaCl dm 3 Preparation

250 ml 0.5 mole NaCl dm 3 Preparation What units will we have when the problem is worked?

250 ml 0.5 mole NaCl dm 3 Preparation need grams

250 ml 0.5 mole NaCl dm 3 Preparation need grams

250 ml 0.5 mole NaCl dm 3 1 mole NaCl Preparation need grams

250 ml 0.5 mole NaCl 58 g NaCl dm 3 1 mole NaCl Preparation need grams

250 ml 0.5 mole NaCl 58 g NaCl dm 3 1 mole NaCl Preparation Now What? need grams

250 ml 0.5 mole NaCl 58 g NaCl 1 dm 3 dm 3 1 mole NaCl 1000 ml Preparation need grams

250 ml 0.5 mole NaCl 58 g NaCl 1 dm 3 dm 3 1 mole NaCl 1000 ml Preparation 7.3 grams NaCl

Practice Problems

0.37 g CaCl 2 1 mole 1000 ml 340 ml 110 g 1 dm M CaCl 2 Practice Problem 1

50 cm mole Al(OH) 3 78 g Al(OH) 3 1 dm 3 dm 3 mole 1000 cm g Al(OH) 3 Practice Problem 2

Homework Problems

50 g NaOH 1 mole 1000 cm cm 3 40 g 1 dm M NaOH Homework Problem 1

100 cm mole CaSO g 1 dm 3 d m 3 mole 1000 ml 3.4 g CaSO 4 Homework Problem 2

100 ml 0.5 mole HCl 36 g 1 dm 3 dm 3 mole 1000 ml 1.8 g HCl Homework Problem 3

Making Dilutions

A solution can be made less concentrated by dilution with solvent Making Dilutions

M 1 V 1 = M 2 V 2 original solution 1 = diluted solution 2 Volume units must be the same for both volumes in this equation. Making Dilutions

How do you prepare 100ml of 0.40M MgSO 4 from a stock solution of 2.0M MgSO 4 ? Dilution Problem

M 1 V 1 = M 2 V 2 M 1 = V 1 = M 2 = V 2 = Dilution Problem

M 1 V 1 = M 2 V 2 M 1 = 2.0M MgSO 4 V 1 = unknown M 2 = 0.40M MgSO 4 V 2 = 100ml Dilution Problem

Step 1 - write the equation: M 1 V 1 = M 2 V 2 Step 2 - manipulate the equation: V 1 = M 2 V 2 / M 1 Step 3 - add the numbers: V 1 = (0.40M) (100ml) / 2.0M Dilution Problem

Step 4 - do the calculation: V 1 = 20ml Step 5 - describe the preparation: Add 80ml of distilled water to 20ml of the 2.0 M MgSO 4 solution Dilution Problem

Homework Problems

Molarity Calculations: M (NH 4 ) 2 C 4 H 4 O M CoSO M Fe(NO 3 ) 2

Preparations: 1.390g NiCl g AgF 3.Use 50cm 3 of the 1.0M NaCl solution. Add 200cm 3 of distilled water to make the total volume 250cm 3.

Dilutions: 1.Add 19.2cm 3 of 0.52M CoCl 2 solution to a graduate. Add distilled water to make the total volume 100cm 3. 2.Add 72.5cm 3 of 0.69M Ba(NO 3 ) 2 solution to a graduate. Add distilled water to make the total volume 200cm 3. 3.Add 37.5ml of 2M NH 4 Br solution to a graduate. Add distilled water to make the total volume 500ml.

Solution Preparation

Other Solution Concentrations

Molality moles solute m = Kg solvent

Normality equivalents equivalents solute N = dm 3 solution