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Taylor Walsh Shiv Patel Emily Penn Philip Adejumo

Acid-Base Titrations. Chapter 17-4. Taylor Walsh Shiv Patel Emily Penn Philip Adejumo. Objectives. Be able to read a titration curve Understand how titrations work Perform titration calculations. Introduction.

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Taylor Walsh Shiv Patel Emily Penn Philip Adejumo

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  1. Acid-Base Titrations Chapter 17-4 Taylor Walsh Shiv Patel Emily Penn Philip Adejumo

  2. Objectives • Be able to read a titration curve • Understand how titrations work • Perform titration calculations

  3. 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

  4. 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

  5. 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 base • Weak acid-strong base • Polyprotic acid-strong base

  6. Final pH Rapid Rise Portion Initial pH Equivalence Point http://www.files.chem.vt.edu/chem-ed/titration/graphics/titration-strong-acid-35ml.gif

  7. http://0.tqn.com/d/chemistry/1/0/f/g/satitration.JPG Strong Acid-Strong Base • The initial pH • The initial pH is a purely acidic solution • Between the initial pH and the equivalence point • pH slowly rises at first, then more Rapidly when it gets close to the Equivalence point • The equivalence point • After the equivalence point 4 3 2 1 Ex. .100 M NaOH added to 50.0 mL of .100 M HCl

  8. Calculating pH for a Strong Acid-Strong Base Titration • First determine how many moles of H+ were originally present and how many moles of OH- were added • 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

  9. Ex. Calculate the pH of 49.0 mL of 0.100 M NaOH solution after 50.0 mL of 0.100 M HClsolution was added • (0.0500L soln)( )= 5.00 x 10-3 mol H+ • (0.0490 L soln)()= 4.90 x 10-3mol OH- • (5.00 x 10-3 mol H+) – (4.90 x 10-3mol OH- ) = 0.10 x 10-3mol H+ • [H+] = = = = 1.0 x 10-3 • pH= -log (1.0 x 10-3) = 3.00

  10. Weak-Acid Strong Base Titrations 4 • The initial pH • pH of the acid • Between the initial pH and the Equivalence point • The equivalence point • After the equivalence point 3 2 1

  11. 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

  12. 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)

  13. (.0500 L soln)()= 5.00 x 10-3 mol HC2H3O2 (.0450 L soln)() = 4.50 x 10-3 molNaOH

  14. 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

  15. 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

  16. 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

  17. Kb= = = 5.6 x 10-10 X = [OH-] = 5.3 x 10-6 M pOH = 5.28 pH = 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

  18. Polyprotic Acids • When weak acids contain more than one ionizable H atom (H3PO3) • Neutralization occurs in a series of steps H3PO3 HPO3-2 H2PO3- http://www.files.chem.vt.edu/RVGS/APChem/lab/Experiments/images/titration_curve.jpg

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