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Quantitative Aspects of Reactions in Solution Sections 4.5-4.7. Terminology. In solution we need to define the SOLVENT the component whose physical state is preserved when solution forms; usually the component in the largest proportion - PowerPoint PPT Presentation
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© 2006 Brooks/Cole - Thomson
Quantitative Aspects of Reactions in
SolutionSections 4.5-4.7
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© 2006 Brooks/Cole - Thomson
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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© 2006 Brooks/Cole - Thomson
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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© 2006 Brooks/Cole - Thomson
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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© 2006 Brooks/Cole - Thomson
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
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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PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
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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© 2006 Brooks/Cole - Thomson
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
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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© 2006 Brooks/Cole - Thomson
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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© 2006 Brooks/Cole - Thomson
[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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© 2006 Brooks/Cole - Thomson
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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© 2006 Brooks/Cole - Thomson
• 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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© 2006 Brooks/Cole - Thomson
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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© 2006 Brooks/Cole - Thomson
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?