Ch. 13: SolutionsCh. 13: Solutions

Dr. Namphol Sinkaset

Chem 152: Introduction to General Chemistry

I. Chapter OutlineI. Chapter Outline

I. IntroductionII. ConcentrationIII. Preparation of a SolutionIV. DilutionV. Solution Stoichiometry

I. IntroductionI. Introduction

• Solution chemistry is the most well studied – why?

• Although solutions tend to be liquid based, the general definition allows for other types of solutions.

• solution: homogeneous mixture of two or more substances

I. Types of SolutionsI. Types of Solutions

I. Solution ComponentsI. Solution Components• There are two parts of a solution.

solute: substance present in smaller amount

solvent: substance present in larger amount

• For stoichiometry, the important aspect of a solution is its concentration.

• concentration: amount of solute present in a certain volume of solution

II. Solution ConcentrationII. Solution Concentration• Solutions can be either dilute or

concentrated.• dilute: small amount of solute relative to

amount of solvent• concentrated: large amount of solute

relative to amount of solvent• There are several different ways to

express solution concentration.

II. Mass PercentII. Mass Percent

• Mass percent expresses the number of grams of solute per 100 g of solution.

%100solvent mass solute mass

solute mass percent mass

II. Using Mass PercentII. Using Mass Percent

• If expressed as a fraction over 100, the mass percent can be used as a conversion factor.

• For a solution that is 23.2% ethanol by mass:

solution g 100.0ethanol g 23.2

ethanol g 23.2solution g 100.0

OR

II. Sample ProblemII. Sample Problem

• Calculate the mass percent of a solution containing 15.5 g of fructose and 249.6 g of water.

II. Sample ProblemII. Sample Problem

• Ocean water contains 3.5% sodium chloride by mass. How much sodium chloride does a 2.00 L sample of ocean water contain? Note that ocean water has a density of 1.027 g/mL.

II. MolarityII. Molarity

• The most common concentration unit is molarity, which is moles solute per L of solution.

In a solution, solute is evenly dispersed in the solvent!!

II. Using MolarityII. Using Molarity

• Molarity can be used as a conversion factor between moles of solute and liters of solution.

• For a 0.500 M NaCl solution:

solution L 1NaCl mole 0.500

NaCl mole 0.500solution L 1

OR

II. Sample ProblemII. Sample Problem

• Calculate the molarity of a solution formed when 24.2 g NaCl is dissolved in 124.1 mL of water.

II. Sample ProblemII. Sample Problem

• How many grams of Na2HPO4 are needed to make 1.50 L of a 0.500 M Na2HPO4 solution?

II. Concentration of IonsII. Concentration of Ions

• When soluble ionic compounds are added to water, they dissociate.

• Soluble ionics dissociate because the solvent-solute attraction is greater than the solute-solute attraction.

II. The Dissolving ProcessII. The Dissolving Process

II. NaCl in SolutionII. NaCl in Solution

II. Ion ConcentrationsII. Ion Concentrations• Sometimes, we need to know the

concentration of an individual ion.• When calculating, we must account for

the ratio seen in the formula of the ionic compound.

• e.g. MgCl2 has two anions for every one cation; anion will be twice as concentrated.

II. Sample ProblemII. Sample Problem

• Calculate the concentration of the ions when 19.6 g of iron(III) sulfate is dissolved in enough water to make 200.0 mL of solution.

III. Solution CreationIII. Solution Creation• The last sample problem is an example

of a calculation needed in order to create a certain volume of solution of a certain concentration.

• This type of calculation is very common in any research lab.

• To make the solution, special glassware and a specific procedure must be used.

III. Creating 1.00 M NaClIII. Creating 1.00 M NaCl

IV. DilutionIV. Dilution

• Less concentrated solutions can be made from more concentrated solutions in a process called dilution.

• The more concentrated solution is known as a stock solution.

• To perform a dilution, you need to know how much of the stock solution to use.

IV. Dilution EquationIV. Dilution Equation

• The following dilution equation makes it easy to calculate how much stock solution is needed.

2211 VMVM

IV. Sample ProblemIV. Sample Problem

• How many mL of a 2.0 M NaCl solution are needed to make 250.0 mL of a 0.50 M NaCl solution?

V. Solution StoichiometryV. Solution Stoichiometry

• Since molarity is a ratio between moles of solute and volume of solution, it can be used in stoichiometric calculations.

• The key is to remember that molarity breaks down into units of mole/L!

V. Sample ProblemV. Sample Problem

• How many mL of 0.10 M HCl reacts with 0.10 g Al(OH)3 according to the reaction below?

Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

V. Sample ProblemV. Sample Problem

• How much PbCl2 forms when 267 mL 1.50 M lead(II) acetate reacts with 125 mL 3.40 M sodium chloride according to the reaction below?

Pb(CH3COO)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaCH3COO(aq)