Ch. 13: Solutions Dr. Namphol Sinkaset Chem 152: Introduction to General Chemistry

I. Chapter Outline I.Introduction II.Concentration III.Preparation of a Solution IV.Dilution V.Solution Stoichiometry

I. 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 Solutions

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

II. 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: OR

II. Sample Problem Calculate the mass percent of a solution containing 15.5 g of fructose and g of water.

II. 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 g/mL.

II. 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 Molarity Molarity can be used as a conversion factor between moles of solute and liters of solution. For a M NaCl solution: OR

II. Sample Problem Calculate the molarity of a solution formed when 24.2 g NaCl is dissolved in mL of water.

II. Sample Problem How many grams of Na 2 HPO 4 are needed to make 1.50 L of a M Na 2 HPO 4 solution?

II. 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 Process

II. NaCl in Solution

II. 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. MgCl 2 has two anions for every one cation; anion will be twice as concentrated.

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

III. 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 NaCl

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

IV. Sample Problem How many mL of a 2.0 M NaCl solution are needed to make mL of a 0.50 M NaCl solution?

V. 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 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)  AlCl 3(aq) + 3H 2 O (l)

V. Sample Problem How much PbCl 2 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(CH 3 COO) 2(aq) + 2NaCl (aq)  PbCl 2(s) + 2NaCH 3 COO (aq)