Taylor Walsh Shiv Patel Emily Penn Philip Adejumo

Taylor WalshShiv Patel

Emily PennPhilip Adejumo

ACID-BASE TITRATIONSChapter 17-4

OBJECTIVES Be able to read a titration curve Understand how titrations work Perform titration calculations

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

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

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

Equivalence Point

Rapid Rise Portion

Initial pH

Final pH

http://www.files.chem.vt.edu/chem-ed/titration/graphics/titration-strong-acid-35ml.gif

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

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http://0.tqn.com/d/chemistry/1/0/f/g/satitration.JPG

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

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

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

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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

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)

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

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

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

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

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

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