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Taylor WalshShiv Patel
Emily PennPhilip Adejumo
ACID-BASE TITRATIONSChapter 17-4
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OBJECTIVES Be able to read a titration curve Understand how titrations work Perform titration calculations
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INTRODUCTION Equivalence point- the point at which
stoichiometrically equivalent quantities of acids and bases have been brought together
http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
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INTRODUCTION (CONT’D) Titration- when a solution containing a
known concentration of base is slowly added to an acid (or vice versa)Titration enables us to find the equivalence
point of the acid-base solution
http://www.dartmouth.edu/~chemlab/techniques/graphics/titration/titration6.gif
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TITRATION CURVES A titration curve is a graph of the pH
as a function of the volume of the added acid or base
There are 3 types of titrations with distinct titration curves:Strong acid-strong baseWeak acid-strong basePolyprotic acid-strong base
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Equivalence Point
Rapid Rise Portion
Initial pH
Final pH
http://www.files.chem.vt.edu/chem-ed/titration/graphics/titration-strong-acid-35ml.gif
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STRONG ACID-STRONG BASE1. The initial pH
a. The initial pH is a purely acidicsolution
2. Between the initial pH and the equivalence point
pH slowly rises at first, then moreRapidly when it gets close to theEquivalence point
3. The equivalence point4. After the equivalence point
Ex. .100 M NaOH added to 50.0 mL of .100 M HCl
12
3
4
http://0.tqn.com/d/chemistry/1/0/f/g/satitration.JPG
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CALCULATING PH FOR A STRONG ACID-STRONG BASE TITRATION1. First determine how many moles of H+ were
originally present and how many moles of OH- were added
2. Subtract the two values (moles) to calculate moles of H+
There are more moles of H+ than moles of OH-, so the resulting value will be moles of H+Ex. Calculate the pH when the
following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution
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Ex. Calculate the pH of 49.0 mL of 0.100 M NaOH solution after 50.0 mL of 0.100 M HCl solution was added
(0.0500L soln)( )= 5.00 x 10-3 mol H+
(0.0490 L soln)()= 4.90 x 10-3 mol OH-
(5.00 x 10-3 mol H+) – (4.90 x 10-3 mol OH- ) = 0.10 x 10-3 mol H+
[H+] = = = = 1.0 x 10-3
pH= -log (1.0 x 10-3) = 3.00
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WEAK-ACID STRONG BASE TITRATIONS1. The initial pH
pH of the acid
2. Between the initial pH and the
Equivalence point
3. The equivalence point
4. After the equivalence point
12
3
4
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STRONG ACID- STRONG BASE VS. WEAK ACID-STRONG BASE The solution of the weak acid has a
higher initial pH than a solution of a strong acid of the same concentration
The pH change at the rapid-rise portion of the curve is smaller for the weak acid than it is for the strong acid
The pH at the equivalence point is above 7.00 for the weak acid-strong base titration Equivalence point for strong acid-strong base
is always at 7.00 pH
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CALCULATING PH FOR A WEAK ACID-STRONG BASE TITRATION Calculate [HX] and [X-] after reaction Use [HX], [X-], and Ka to calculate Use [H+] to calculate pH
Ex: Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC2H3O2, (Ka = 1.8 x 10-5)
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Before rxn (5.00 x mol) (4.50 x mol) 0.0 mol
HC2H3O2 (aq) + OH- C2H3O2-
(aq)After rxn (0.50 x ) 0.0 mol 4.50 x mol
(.0500 L soln)()= 5.00 x 10-3 mol HC2H3O2
(.0450 L soln)() = 4.50 x 10-3 mol NaOH
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45.0 mL + 50.0 mL = 0.0950 L
[HC2H3O2] = = .0053 M
[C2H3O2-] = = .0474 M
Ka = = 1.8 x 10-5
[H+] = Ka x = (1.8 x 10-5) x = 2.0 x 10-6 M
pH = -log(2.0 x 10-6) = 5.70
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EXTRA QUESTION Calculate the pH at the equivalence
point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
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CALCULATE THE PH AT THE EQUIVALENCE POINT IN THE TITRATION OF 50.0 ML OF 0.100 M HC2H3O2 WITH 0.100 M NAOH
Moles=M x L= (0.100 mol/L)(0.0500 L) = 5.00 x 10-3 mol HC2H3O2
[C2H3O2-]= = (0.0500 M)
Since C2H3O2- is a weak base:
C2H3O2- (aq) + H2O (l) HC2H3O2 (aq) +
OH- (aq)Kb= = = 5.6x10-10
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Kb= = = 5.6 x 10-10
X = [OH-] = 5.3 x 10-6 MpOH = 5.28pH = 8.72
Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
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POLYPROTIC ACIDS When weak acids contain more than one
ionizable H atom (H3PO3) Neutralization occurs in a series of steps
H3PO3
H2PO3-
HPO3-2
http://www.files.chem.vt.edu/RVGS/APChem/lab/Experiments/images/titration_curve.jpg