Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 12.5 Molarity and Dilution Chapter 12...

Basic Chemistry Copyright © 2011 Pearson Education, Inc.1

12.5

Molarity and Dilution

Chapter 12 Solutions


Molarity (M)Molarity (M) is

• a concentration term for solutions

• the moles of solute in 1 L of solution

• moles of soluteliter of solution


Preparing a 6.0 M Solution

A 6.00 M NaOH solution is prepared

• by weighing out 60.0 g of NaOH (1.50 mol) and

• adding water to make 0.250 L of a 6.00 MNaOH solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Preparation of Solutions


A solution of a desired concentration can be prepared by diluting a small volume of a more-concentrated solution, a stock solution, with additional solvent.

– Calculate the number of moles of solute desired in the final volume of the more-dilute solution and then calculate

the volume of the stock solution that contains the amount of solute.

– Diluting a given quantity of stock solution with solvent does not change the number of moles of solute present.

– The relationship between the volume and concentration of the stock solution and the volume and concentration of the

desired diluted solution is

(Vs) (M s) = moles of solute = (Vd) (M d).



Calculating Molarity


What is the molarity of 0.500 L of a NaOH solution if it contains 6.00 g of NaOH?STEP 1 State the given and needed quantities.

Given 6.00 g of NaOH in 0.500 L of solution Need molarity (M) STEP 2 Write a plan to calculate molarity.

molarity (M) = moles of solute liters of solution

grams of NaOH moles of NaOH molarity

Example of Calculating Molarity


STEP 3 Write equalities and conversion factors needed.

1 mol of NaOH = 40.01 g of NaOH 1 mol NaOH and 40.01 g NaOH

40.01 g NaOH 1 mol NaOHSTEP 4 Set up problem to calculate molarity.6.00 g NaOH x 1 mol NaOH = 0.150 mol of NaOH 40.01 g NaOH 0.150 mol NaOH = 0.300 mol0.500 L solution 1 L

= 0.300 M NaOH solution

Example of Calculating of Molarity (continued)


What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?

A. 0.557 M NaHCO3 solution

B. 1.44 M NaHCO3 solution

C. 1.71 M NaHCO3 solution

Learning Check


STEP 1 State the given and needed quantities.

Given 46.8 g of NaHCO3 in 0.325 L of solution

Need molarity (M)

STEP 2 Write a plan to calculate molarity.

molarity (M) = moles of solute

liters of solution

grams of NaHCO3 moles of NaHCO3 molarity

Solution



1 mol of NaHCO3 = 84.01 g of NaHCO3

1 mol NaHCO3 and 84.01 g NaHCO3

84.01 g NaOH 1 mol NaHCO3

STEP 4 Set up problem to calculate molarity.

46.8 g NaHCO3 x 1 mol NaHCO3 = 0.557 mol of NaHCO3

84.01 g NaHCO3

0.557 mol NaHCO3 = 1.71 M NaHCO3 solution

0.325 L

Solution (continued)


Molarity Conversion Factors

The units of molarity are used as conversion factors in calculations with solutions.


Example of Using Molarity in Calculations

How many grams of KCl are needed to prepare 0.125 L of a 0.720 M KCl solution?STEP 1 State the given and needed quantities.

Given 0.125 L of a 0.720 M KCl solution Need grams of KCl

STEP 2 Write a plan to calculate mass or volume.

liters of KCl solution moles of KCl grams of KCl


Example of Using Molarity in Calculations (continued)


1 mol of KCl = 74.55 g of KCl 1 mol KCl and 74.55 g KCl

74.55 g KCl 1 mol KCl 1 L of KCl solution = 0.720 mol of KCl

1 L KCl solution and 0.720 mol KCl 0.720 mol KCl 1 L KCl solution


Example of Using Molarity in Calculations (continued)

STEP 4 Set up problem to calculate mass or volume.0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g of KCl 1 L 1 mol KCl


How many grams of AlCl3 are needed to prepare 37.8 mL of a 0.150 M AlCl3 solution?

A. 0.00567 g of AlCl3

B. 0.756 g of AlCl3

C. 5.04 g of AlCl3

Learning Check


Solution


Given 37.8 mL of a 0.150 M AlCl3 solution

Need grams of AlCl3


milliliters of AlCl3 solution liters of AlCl3 solution

moles of AlCl3 grams of AlCl3




1 mol of AlCl3 = 133.33 g of AlCl3

1 mol AlCl3 and 133.33 g AlCl3 133.33 g AlCl3 1 mol AlCl3

1000 mL of AlCl3 solution = 1 L of AlCl3 solution

1000 mL AlCl3 solution and 1 L AlCl3 solution 1 L AlCl3 solution 1000 mL AlCl3 solution

1 L of AlCl3 solution = 0.150 mol of AlCl3

1 L AlCl3 solution and 0.150 mol AlCl3 0.150 mol AlCl3 1 L AlCl3 solution



STEP 4 Set up problem to calculate mass or volume.

37.8 mL x 1 L x 0.150 mol x 133.33 g 1000 mL 1 L 1 mol

= 0.756 g of AlCl3 (B)


How many milliliters of a 2.00 M HNO3 solutioncontain 24.0 g of HNO3?

A. 12.0 mL of HNO3 solution

B. 83.3 mL of HNO3 solution

C. 190. mL of HNO3 solution

Learning Check



Given 24.0 g of HNO3; 2.00 M HNO3 solution

Need milliliters of HNO3 solution


g of solution moles of HNO3 mL of HNO3

Solution



1 mol of HNO3 = 63.02 g of HNO3

1 mol HNO3 and 63.02 g HNO3

63.02 g HNO3 1 mol HNO3

1000 mL of HNO3 = 2.00 mol of HNO3

1000 mL HNO3 and 2.00 mol HNO3

2.00 mol HNO3 1000 mL HNO3



STEP 4 Set up problem to calculate mass or volume.

24.0 g HNO3 x 1 mol HNO3 x 1000 mL

63.02 g HNO3 2.00 mol HNO3

= 190. mL of HNO3 solution (C)



DilutionIn a dilution,

• water is added

• volume increases

• concentration decreases


Na+

NaNO3 solution

NO3-

Moles = 1.0

Volume = 1.0 L

Molarity = 1.0 M


• Solution volume is doubled

• Solution concentration is halved

• Moles of solute remain the same

Na+

NO3-

Moles = 1.0

Volume = 2.0 L

Molarity = 0.50 M


Comparing Initial and Diluted Solutions

In the initial and diluted solution,

• the moles of solute are the same

• the concentrations and volumes are related by the equation

M1V1 = M2V2

initial diluted


Calculating Dilution Quantities


Example of Dilution Calculations

What is the final molarity of the solution when 0.180 L of 0.600 M KOH is diluted to 0.540 L?

STEP 1 Prepare a table of the initial and diluted volumes and concentrations.

Initial Solution Diluted Solution

M1 = 0.600 M M2 = ?

V1 = 0.180 L V2 = 0.540 L


Example of Dilution Calculations (continued)STEP 2 Solve the dilution expression for the

unknown quantity. M1V1 = M2V2

M1V1 = M2V2

V2 V2

M2 = M1V1

V2

STEP 3 Set up the problem by placing known quantities in the dilution expression.

M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 MV2 0.540 L


Learning Check

What is the final volume, in milliliters, if 15.0 mL of a 1.80 M KOH solution is diluted to give a 0.300 M KOH solution?

A. 27.0 mL of 0.300 M KOH solution

B. 60.0 mL of 0.300 M KOH solution

C. 90.0 mL of 0.300 M KOH solution


Solution

STEP 1 Prepare a table of the initial and diluted volumes and concentrations.Initial Solution Diluted Solution M1= 1.80 M V1 = 15.0 mL M2= 0.300 M V2 = ?



STEP 2 Solve the dilution expression for the unknown quantity. M1V1 = M2V2

M1V1 = M2V2

M2 M2

V2 = M1V1

M2

STEP 3 Set up the problem by placing known quantities in the dilution expression.

V2 = M1V1 = (1.80 M)(15.0 mL) M2 0.300 M = 90.0 mL (C )

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 12.5 Molarity and Dilution Chapter 12...

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