1 © 2006 brooks/cole - thomson quantitative aspects of reactions in solution sections 5.8-5.10

1

© 2006 Brooks/Cole - Thomson

Quantitative Quantitative Aspects of Aspects of Reactions in Reactions in

SolutionSolutionSections 5.8-5.10Sections 5.8-5.10

2


TerminologyTerminologyTerminologyTerminologyIn solution we need to define the In solution we need to define the • SOLVENTSOLVENT the component whose the component whose

physical state is physical state is preserved when preserved when solution forms; usually the solution forms; usually the component in the largest component in the largest proportionproportion

• SOLUTESOLUTEthe other solution componentthe other solution component

3


Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a solution The amount of solute in a solution is given by its is given by its concentrationconcentration.

Molarity(M) = moles solute

liters of solutionConcentration (M) = [ …]

4


1.0 L of water 1.0 L of water was used to was used to

make 1.0 L of make 1.0 L of solution. Notice solution. Notice

the water left the water left over.over.

CCR, page 206

5


Preparing a SolutionPreparing a Solution

Active Figure 5.18

6


Preparing SolutionsPreparing SolutionsPreparing SolutionsPreparing Solutions

• Weigh out a solid Weigh out a solid solute and dissolve in a solute and dissolve in a given quantity of given quantity of solvent.solvent.

• DiluteDilute a concentrated a concentrated solution to give one solution to give one that is less that is less concentrated.concentrated.

7


PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 •6 HH22O in enough water to make 250 mL O in enough water to make 250 mL of solution. Calculate molarity.of solution. Calculate molarity.

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 •6 HH22O in enough water to make 250 mL O in enough water to make 250 mL of solution. Calculate molarity.of solution. Calculate molarity.

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22•6H•6H22OO

5.00 g • 1 mol

237.7 g = 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Step 2: Calculate molarityCalculate molarity

[NiClNiCl22•6 H•6 H22OO ] = 0.0841 M

8


The Nature of a CuClThe Nature of a CuCl22 Solution: Solution:Ion ConcentrationsIon Concentrations

CuClCuCl22(aq) --> (aq) -->

CuCu2+2+(aq) + (aq) + 22 Cl Cl--(aq)(aq)

If [CuClIf [CuCl22] = 0.30 M, ] = 0.30 M,

then then

[Cu[Cu2+2+] = 0.30 M] = 0.30 M

[Cl[Cl--] = 2 x 0.30 M] = 2 x 0.30 M

9


Step 1: Step 1: Calculate moles of acid required.Calculate moles of acid required.

(0.0500 mol/L)(0.250 L) = 0.0125 mol(0.0500 mol/L)(0.250 L) = 0.0125 mol

Step 2: Step 2: Calculate mass of acid required.Calculate mass of acid required.

(0.0125 mol )(90.00 g/mol) = (0.0125 mol )(90.00 g/mol) = 1.13 g1.13 g

USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

moles = M•Vmoles = M•V

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, is

required to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?

10


Preparing a Solution by Preparing a Solution by DilutionDilution

Preparing a Solution by Preparing a Solution by DilutionDilution

11


PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you do?What do you do?


3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

Add water to the 3.0 M solution to Add water to the 3.0 M solution to lower its concentration to 0.50 M lower its concentration to 0.50 M

Dilute the solution!Dilute the solution!

12


PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??

PROBLEM: You have 50.0 mL of 3.0 M PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. NaOH and you want 0.50 M NaOH. What do you doWhat do you do??

How much water is added?How much water is added?

The important point is that --->The important point is that --->

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

13


PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?

Amount of NaOH in original solution = Amount of NaOH in original solution =

M • VM • V = =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = Amount of NaOH in final solution must also = 0.15 mol NaOH0.15 mol NaOH

Volume of final solution =Volume of final solution =

(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L

or or 300 mL300 mL

14




Conclusion:Conclusion:

add 250 mL add 250 mL of waterof water to to 50.0 mL of 50.0 mL of 3.0 M NaOH 3.0 M NaOH to make 300 to make 300 mL of 0.50 M mL of 0.50 M NaOH. NaOH.

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

15


A shortcutA shortcut

CCinitialinitial • V • Vinitialinitial = C = Cfinalfinal • V • Vfinalfinal

Preparing Solutions Preparing Solutions by Dilutionby Dilution

16


pH, a Concentration pH, a Concentration ScaleScale

pH: a way to express acidity -- the concentration of pH: a way to express acidity -- the concentration of HH++ in solution. in solution.

Low pH: high [HLow pH: high [H++]] High pH: low [HHigh pH: low [H++]]

Acidic solutionAcidic solution pH < 7pH < 7 Neutral Neutral pH = 7pH = 7 Basic solution Basic solution pH > 7pH > 7

Acidic solutionAcidic solution pH < 7pH < 7 Neutral Neutral pH = 7pH = 7 Basic solution Basic solution pH > 7pH > 7

17


The pH ScaleThe pH Scale

pH pH = log (1/ [H= log (1/ [H++]) ])

= - log [H= - log [H++]]In a In a neutralneutral solution, solution,

[H[H++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M at 25 M at 25 ooCC

pH = - log [HpH = - log [H++] = -log (1.00 x 10] = -log (1.00 x 10-7-7) ) = - [0 + (-7)] = - [0 + (-7)]

= 7= 7

18


[H[H++] and pH] and pHIf the [HIf the [H++] of soda is 1.6 x 10] of soda is 1.6 x 10-3-3 M, M,

the pH is ____?the pH is ____?

Because pH = - log [HBecause pH = - log [H++] ]

thenthen

pH= - log (1.6 x 10pH= - log (1.6 x 10-3-3) )

pH = -{log (1.6) + log (10pH = -{log (1.6) + log (10-3-3)})}

pH = -{0.20 - 3.00)pH = -{0.20 - 3.00)

pH = 2.80pH = 2.80

19


pH and [HpH and [H++]]

If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.

Because pH = - log [HBecause pH = - log [H++] then] then

log [Hlog [H++] = - pH] = - pH

Take antilog and getTake antilog and get

[H[H++] = 10] = 10-pH-pH

[H[H++] = 10] = 10-3.12-3.12 = = 7.6 x 7.6 x 1010-4-4 M M

20


• Zinc reacts with Zinc reacts with acids to produce Hacids to produce H22 gas. gas.

• Have 10.0 g of ZnHave 10.0 g of Zn

• What volume of What volume of 2.50 M HCl is 2.50 M HCl is needed to convert needed to convert the Zn completely?the Zn completely?

SOLUTION SOLUTION STOICHIOMETRYSTOICHIOMETRY

Section 5.10Section 5.10

21


GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY CALCULATIONSSTOICHIOMETRY CALCULATIONS

Mass zinc

StoichiometricfactorMoles

zincMoles HCl

Mass HCl

VolumeHCl

22


Step 1: Step 1: Write the balanced equationWrite the balanced equation

Zn(s) + 2 HCl(aq) --> ZnClZn(s) + 2 HCl(aq) --> ZnCl22(aq) + H(aq) + H22(g)(g)

Step 2: Step 2: Calculate amount of ZnCalculate amount of Zn

10.0 g Zn • 1.00 mol Zn65.39 g Zn

= 0.153 mol Zn

Zinc reacts with acids to produce HZinc reacts with acids to produce H22 gas. If you gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?needed to convert the Zn completely?


Step 3: Step 3: Use the stoichiometric factorUse the stoichiometric factor

23


Step 3: Step 3: Use the stoichiometric factorUse the stoichiometric factor

0.153 mol Zn • 2 mol HCl1 mol Zn

= 0.306 mol HCl

0.306 mol HCl • 1.00 L

2.50 mol = 0.122 L HCl



Step 4: Step 4: Calculate volume of HCl req’dCalculate volume of HCl req’d

24


ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->

acidacid basebase

NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

Carry out this reaction using a Carry out this reaction using a TITRATIONTITRATION..

Oxalic acid,Oxalic acid,

HH22CC22OO44

25


Setup for titrating an acid with a baseSetup for titrating an acid with a base

Active Figure 5.23

26


TitratioTitrationn

1. Rinse buret with titrant. 1. Rinse buret with titrant. Remove bubbles. Add Remove bubbles. Add solution from the buret to solution from the buret to flask. flask.

2. Flask contains analyte 2. Flask contains analyte and indicator. Base reacts and indicator. Base reacts with acid in solution in the with acid in solution in the flask.flask.

3. Indicator shows when 3. Indicator shows when exact stoichiometric exact stoichiometric reaction has occurred.reaction has occurred.

4. Net ionic equation4. Net ionic equation

HH++ + OH + OH-- --> H --> H22OO

5. At equivalence point (end 5. At equivalence point (end point)point)

moles Hmoles H++ = moles OH = moles OH--

27


1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires (oxalic acid) requires

35.62 mL of NaOH for titration to an 35.62 mL of NaOH for titration to an

equivalence point. What is the equivalence point. What is the

concentration of the NaOH?concentration of the NaOH?

LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.

LAB PROBLEM #1: Standardize a LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately solution of NaOH — i.e., accurately determine its concentration.determine its concentration.

28


1.065 g of H1.065 g of H22CC22OO44 (oxalic acid) requires 35.62 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?point. What is the concentration of the NaOH?


Step 1: Step 1: Calculate amount of HCalculate amount of H22CC22OO44

1.065 g • 1 mol

90.04 g = 0.0118 mol

0.0118 mol acid • 2 mol NaOH1 mol acid

= 0.0236 mol NaOH

Step 2: Step 2: Calculate amount of NaOH req’dCalculate amount of NaOH req’d

29




Step 1: Step 1: Calculate amountCalculate amount of Hof H22CC22OO44

= 0.0118 mol acid= 0.0118 mol acid

Step 2: Step 2: Calculate amount of NaOH req’dCalculate amount of NaOH req’d

= 0.0236 mol NaOH= 0.0236 mol NaOH

Step 3: Step 3: Calculate concentration of NaOHCalculate concentration of NaOH

0.0236 mol NaOH0.03562 L

0.663 M

[NaOH] = 0.663 M[NaOH] = 0.663 M

1 © 2006 brooks/cole - thomson quantitative aspects of reactions in solution sections 5.8-5.10

Documents