Quantitative Aspects of Reactions in Solution Sections 4.5-4.7

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© 2006 Brooks/Cole - Thomson

Quantitative Aspects of Reactions in

SolutionSections 4.5-4.7

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TerminologyIn solution we need to define the • SOLVENT the component whose

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

• SOLUTEthe other solution component

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Concentration of Solute

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

Concentration (M) = [ …]

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The Nature of a CuCl2 Solution:Ion Concentrations

CuCl2(aq) -->

Cu2+(aq) + 2 Cl-(aq)

If [CuCl2] = 0.30 M,

then

[Cu2+] = 0.30 M

[Cl-] = 2 x 0.30 M

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Preparing a Solution

1. Weigh out mass of solute.

2. Dissolve it in a little water in a volumetric flask

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1.0 L of water was used to

make 1.0 L of solution. Notice

the water left over.

CCR, page 206

Do NOT add 1.0 L of water!!!

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PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity.

Step 1: Calculate moles of NiCl2•6H2O

5.00 g • 1 mol

237.7 g = 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Calculate molarity

[NiCl2•6 H2O ] = 0.0841 M

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Step 1: Calculate moles of acid required.

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

Step 2: Calculate mass of acid required.

(0.0125 mol )(90.00 g/mol) = 1.13 g

USING MOLARITY

moles = M•V

What mass of oxalic acid, H2C2O4, is

required to make 250. mL of a 0.0500 Msolution?

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Preparing Solutions

• Weigh out a solid solute and dissolve to a given quantity of solution.

OR

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

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

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water do we add?

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

Dilute the solution!

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How much water is added?

The important point is that --->

MOLES OF NaOH IN ORIGINAL SOLUTION = MOLES OF NaOH IN FINAL SOLUTION

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Amount of NaOH in original solution =

M • V =

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

Amount of NaOH in final solution must also = 0.15 mol NaOH

Volume of final solution =

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

or 300 mL

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Conclusion:

add enough water to the 50.0 mL of 3.0 M NaOH to make

300 mL of

0.50 M NaOH.

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

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A shortcut

Cinitial • Vinitial = Cfinal • Vfinal

Preparing Solutions by Dilution

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pH, a Concentration Scale

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

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

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

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The pH Scale

pH = - log [H+] In a neutral solution,

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

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

= - [0 + (-7)]

= 7

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

the pH is ____?

Because pH = - log [H+]

then

pH= - log (1.6 x 10-3)

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

pH = -{0.20 - 3.00)

pH = 2.80

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pH and [H+]

If the pH of Coke is 3.12, its [H+] is ____________.

Because pH = - log [H+] then

log [H+] = - pH

Take antilog and get

[H+] = 10-pH

[H+] = 10-3.12 = 7.6 x10-4M

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• Zinc reacts with acids to produce H2 gas.

• Have 10.0 g of Zn

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

SOLUTION STOICHIOMETRY

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GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS

Convert something to Moles

CoefficientRatio

Convert to what you’re looking for

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Write the balanced equation

Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)Step 1: Calculate amount of Zn

10.0 g Zn • 1.00 mol Zn65.39 g Zn

= 0.153 mol Zn

Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

Step 2: Use the stoichiometric factor

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Step 2: Use 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

Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

Step 3: Calculate volume of HCl req’d

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ACID-BASE REACTIONSTitrations

H2C2O4(aq) + 2 NaOH(aq) --->

acid base

Na2C2O4(aq) + 2 H2O(liq)

Carry out this reaction using a TITRATION.

OXALIC ACID,

H2C2O4

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Setup for titrating an acid with a base

Active Figure 5.23

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Titration

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

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

3. Indicator shows when exact stoichiometric reaction has occurred.

4. Net ionic equation

H+ + OH- --> H2O5. At equivalence point (end

point) moles H+ = moles OH-

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10.65 mL of 1.2 M H2C2O4 (oxalic acid)

requires 35.62 mL of NaOH for titration

to an equivalence point. What is the

concentration of the NaOH?

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

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10.65 mL of 1.2M H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

Step 1: Calculate amount of H2C2O4

Step 2: Calculate amount of NaOH req’d

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Step 1: Calculate amount of H2C2O4

= 0.01278 mol acid

Step 2: Calculate amount of NaOH req’d

= 0.02556 mol NaOH

Step 3: Calculate concentration of NaOH

[NaOH] = 0.718 M

10.65 mL of 1.2M H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

Quantitative Aspects of Reactions in Solution Sections 4.5-4.7

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