Chemistry: Atoms FirstJulia Burdge & Jason Overby

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Chapter 9 .5

Homework: 61, 63, 65, 67, 71, 73, 75, 77, 79 and 81

Kent L. McCorkle

Cosumnes River College

Sacramento, CA

4.1

A solution is a homogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The solvent is the substance present in the larger amount

Solution Solvent Solute

Soft drink (l)

Air (g)

Soft Solder (s)

H2O

N2

Pb

Sugar, CO2

O2, Ar, CH4

Sn

Solutions

solute

Solvent

When the solvent is water the solution is said to be aqueous

General Properties of Aqueous SolutionsGeneral Properties of Aqueous Solutions

A solution is a homogenous mixture of two or more substances.The substance present in the largest amount (moles) is referred to as the solvent.The other substances present are called the solutes.A substance that dissolves in a particular solvent is said to be soluble in that solvent.

9.1

Concentration of SolutionsConcentration of Solutions

Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.

Other common rearrangements:

moles solutemolarity = liters solution

molL = M

mol = LM

9.5

Worked Example 9.8

Strategy Convert the mass of glucose given to moles, and use the equations for interconversions of M, liters, and moles to calculate the answers.

moles of glucose =

For an aqueous solution of glucose (C6H12O6), determine (a) the molarity of 2.00 L of a solution that contains 50.0 g of glucose, (b) the volume of this solution that would contain 0.250 mole of glucose, and (c) the number of moles of glucose in 0.500 L of this solution.

Solution (a)molarity = = 0.139 M

(b)volume = = 1.80 L

(c)moles of C6H12O6 in 0.500 L = 0.500 L×0.139 M = 0.0695 mol

50.0 g180.2 g/mol

0.277 mol C6H12O6

2.00 L solution 0.250 mol C6H12O6

0.139 M solution

= 0.277 mol

Think About It Check to see that the magnitude of your answers are logical. For example, the mass given in the problem corresponds to 0.277 mole of solute. If you are asked, as in part (b), for the volume that contains a number of moles smaller than 0.277, make sure your answer is smaller than the original volume.

4.5

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

Dilution

Add Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=4.5

How would you prepare 60.0 mL of 0.2 MHNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L

Vi =MfVf

Mi

= 0.200 x 0.064.00

= 0.003 L = 3 mL

3 mL of acid + 57 mL of water = 60 mL of solution

Another Dilution Problem

If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mLWhat would be the resulting concentration?

M1*V1 = M2*V2

(6.5M) * (32 mL) = M2 * (500.0 mL)

M2 = 500 mL6.5 M * 32 mL

M2 = 0.42 M

Concentration of SolutionsConcentration of Solutions

Dilution is the process of preparing a less concentrated solution from a more concentrated one.

moles of solute before dilution = moles of solute after dilution

Concentration of SolutionsConcentration of Solutions

In an experiment, a student needs 250.0 mL of a 0.100 M CuCl2 solution. A stock solution of 2.00 M CuCl2 is available.

How much of the stock solution is needed?

Solution: Use the relationship that moles of solute before dilution = moles of solute after dilution.

(2.00 M CuCl2)(Lc) = (0.100 M CuCl2)(0.2500 L)

Lc = 0.0125 L or 12.5 mLTo make the solution:

1) Pipet 12.5 mL of stock solution into a 250.0 mL volumetric flask.2) Carefully dilute to the calibration mark.

Mc × Lc = Md × Ld

Concentration of SolutionsConcentration of Solutions

Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as

Mc × mLc = Md × mLd

Worked Example 9.9

Strategy Mc = 12.0 M, Md = 0.125 M, mLd = 250.0 mL

What volume of 12.0 M HCl, a common laboratory stock solution, must be used to prepare 150.0 mL of 0.125 M HCl?

Solution 12.0 M × mLc = 0.125 M × 250.0 mL

mLc = 0.125 M × 250.0 mL

12.0 M = 2.60 mL

Think About It Plug the answer into Equation 9.4, and make sure that the product of concentration and volume on both sides of the equation give the same result.

Worked Example 9.10

Strategy (a) Md =

Starting with a 2.0 M stock solution of hydrochloric acid, four standard solutions (1 to 4) are prepared by sequential diluting 10.00 mL of each solution to 250.00 mL. Determine (a) the concentrations of all four standard solutions and (b) the number of moles of HCl in each solution.

Solution (a) Md1 =

Md2 =

Md3 =

Md4 =

2.00 M × 10.00 mL250.00 mL

Mc× mLc

mLd; (b) mol = M×L, 250.00 mL = 2.500×10-1 L

= 8.00×10-2 M

8.00×10-2 M × 10.00 mL250.00 mL = 3.20×10-3 M

3.20×10-3 M × 10.00 mL250.00 mL = 1.28×10-4 M

1.28×10-4 M × 10.00 mL250.00 mL = 5.12×10-6 M

Worked Example 9.10 (cont.)

Think About It Serial dilution is one of the fundamental practices of homeopathy. Some remedies undergo so many serial dilutions that very few (if any) molecules of the original substance still exist in the final preparation.

Solution (b) mol1 = 8.00×10-2 M × 2.500×10-1 L = 2.00×10-2 mol mol2 = 3.20×10-3 M × 2.500×10-1 L = 8.00×10-4 mol mol3 = 1.28×10-4 M × 2.500×10-1 L = 3.20×10-5 mol mol4 = 5.12×10-6 M × 2.500×10-1 L = 1.28×10-6 mol

Worked Example 9.11

Strategy Use the concentration given in each case and the stoichiometry indicated in the corresponding chemical formula to determine the concentration of the specified ion or compound.

Using square-bracket notation, express the concentration of (a) chloride ion in a solution that is 1.02 M in AlCl3, (b) nitrate ion in a solution that is 0.451 M in Ca(NO3)2, and (c) Na2CO3 in a solution in which [Na+] = 0.124 M.

Solution (a) There are 3 moles of Cl- ion for every 1 mole of AlCl3,AlCl3(s) → Al3+(aq) + 3Cl-(aq)

so the concentration of Cl- will be three times the concentration of AlCl3.

[Cl-] = [AlCl3] ×

= ×

= = 3.06 M

3 mol Cl-

1 mol AlCl31.02 mol AlCl3

L3 mol Cl-

1 mol AlCl3

3.06 mol Cl-

L