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Equilibrium for Weak Acids and Bases Practice, Practice, Practice!. Weak Acid Problems What is the pH of the resulting solution when 0.500 moles of acetic acid are dissolved in water and diluted to 1.00 L? Unbalanced reaction:

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Equilibrium for Weak Acids and Bases Practice, Practice, Practice!


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    1. Equilibrium for Weak Acids and BasesPractice, Practice, Practice! Weak Acid Problems What is the pH of the resulting solution when 0.500 moles of acetic acid are dissolved in water and diluted to 1.00 L?Unbalanced reaction: CH3COOH(aq)+H2O(l) ⇄  H3O+(aq) + CH3COO-(aq)     Ka = 1.8x10-5 at 25oC. In a solution of acetic acid, the equilibrium concentrations are found to be [CH3COOH] = 1.000; [CH3COO-] = 0.0042. Find the pH of this solution and the equilibrium constant of ionization of acetic acid. Weak Base Problems What is the pH of the resulting solution when 250.0 mL of a 0.300 M solution of ammonia added to 750.0 mL of water?Unbalanced reaction: NH3(aq) + H2O(l) ⇄   NH4+(aq) + OH-(aq)     Kb = 1.8x10-5 at 25oC. Calculate Kb for methylamine if a solution prepared by dissolving 0.100 moles of methylamine in 1.00 L of water has a measured pH of 11.80.Unbalanced reaction: CH3NH2(aq) + H2O(l)  ⇄  CH3NH3+(aq) + OH-(aq)

    2. [C]c[D]d [C]c[D]d [A]a[B]b [A]a[B]b Equation Summary-Equilibrium Chemical Equilibrium When we have to find x and have a small K and a quadratic (or higher) equation 1000 * Keq << [Initial] When we have BOTH initial products and reactants and we are unsure where we are in reference to equilibrium Ksp = [A+(aq)]a[B-(aq)]b Solubility Equilibrium When we are mixing solutions and are unsure where we are in reference to solubility equilibrium Keq= Q= Qsp = [Ai+(aq)]a[Bi-(aq)]b

    3. [NH4+][OH-] [A-][H3O+] [HA] [NH3] Equation Summary-Equilibrium When we are asked to find pH or given pH and asked to find [H3O+] pH =-log [H30+] When we are asked to find pOH or given pOH and asked to find [OH-] pOH =-log [0H-] Kw= [OH-][H3O+] When we need relate pH and pOH. When we are given one and asked for the other 14 = pH + pOH When we have the concentration of one and we need the concentration of the other Acid-Base Equilibrium When we are mixing solutions and are unsure where we are in reference to solubility equilibrium Kb= Ka=

    4. Relationship Between Ka and Kb • The bigger the Ka of an acid, the smaller the Kb of its conjugate base • The bigger the Kb of a base, the smaller the Ka of its conjugate acid

    5. [A-][H3O+] [HA][OH-] [HA] [A-] How Can We Relate Ka and Kb For Conjugate Acids and Bases? • Acid Reaction HA (aq) + H2O (l)  H3O+ (aq) + A- (aq) • Conjugate Base Reaction A- (aq) + H2O (l)  HA (aq) + OH- (aq) Ka= Kb=

    6. [A-][H3O+] [HA][OH-] [HA] [A-] How Can We Relate Ka and Kb For Conjugate Acids and Bases? • Recall: When we add the 2 reactions together, we multiply their K’s • HA (aq) + H2O (l)  H3O+ (aq) + A- (aq) Ka • A- (aq) + H2O (l)  HA (aq) + OH- (aq) Kb • 2H2O (l)  H3O+(aq) + OH-(aq) Ka*Kb Ka*Kb = x = [H3O+]x[OH-] = Kw

    7. How Can We Relate Ka and Kb For Conjugate Acids and Bases? • Therefore, we can say the following relationship between Ka and Kb exists for CONJUGATE PAIRS • Kw=Ka*Kb • Ka=Kw/Kb • Kb=Kw/Ka • So if only Ka is given, we can find the Kb of its conjugate base when needed • This formula shows that the stronger the weak acid/base, the weaker it’s conjugate base/acid is • Stronger weak acid = weaker conjugate base • Stronger weak base = weaker conjugate acid

    8. How Can We Relate Ka and Kb For Conjugate Acids and Bases?Example • Example 1: The Kb for the rocket fuel hydrazine, N2H4(g) is 1.7 x 10-6. What is the Ka of its conjugate acid N2H5+? • 5.9 x 10-9

    9. Mark received [Ion in Solution] Total Marks Possible [Total Weak Acid] Percent Ionization for Weak AcidsAnother Way of Determining/Reporting [H3O+] • Percent Ionization is • The amount of acid that has dissociated or ionized in solution • We can calculate the percent of weak acid that has ionized in solution in the same manner we calculate the percent we receive on a test… • Percent on test = x 100% • Percent Ionization = x 100%

    10. [Ion in Solution] [H3O+(aq)] [OH-(aq)] [Total Weak Acid/Weak Base in Solution] [HA(aq)] [B(aq)] Percent Ionization Percent Ionization = x 100% Percent Ionization Acid = x 100% Percent Ionization Base= x 100%

    11. Percent Ionization for Weak AcidsExamples • Example 1: The pH of a 0.10mol/L methanoic acid (HCOOH) solution is 2.38. Calculate the percent ionization of methanoic acid • 4.2% • Example 2: Calculate the acid ionization constant, Ka, of acetic acid (CH3COOH) if a 0.1000mol/L solution at equilibrium has a percent ionization of 1.3% • 1.7 x 10-5

    12. Acid-Base Titrations • What is a titration? • A technique used in chemistry used to find unknown concentrations of a sample • Commonly used in industry: • To determine fatty acid content in foods • To test water in aquariums and marine environments • To determine proper concentrations of chemicals used in pharmaceuticals • To determine proper concentrations in biology when using anesthetic or in euthanizing animals

    13. Acid-Base Titrations • How does it work? • We use our knowledge of stoichiometry and acids and bases to determine unknowns from what we do know Acid + Base ⇄ Salt + Water • If we know the concentration of one and have a way to figure out when the reaction finishes, then we can use stoichiometry to figure out the concentration of the other

    14. Acid-Base Titrations • When we perform a titration, the general process is: • We have a solution who’s concentration is known • We react it with a solution who’s concentration is unknown • We use an indicator or pH probe to tell us when the reaction is finished • We figure out the concentration of the unknown using our knowledge of acid-base reactions and stoichiometry

    15. Acid-Base Titrations • So how do we actually do it? • We put our “known solution”-our “titrant” into a buret • We know this solution’s concentration • The buret is graduated so we can measure the volume that we add to the flask • We put our solution with unknown concentration-our analyte-in a flask • With an “indicator” that tells us when the reaction is done by changing colour • We could also use a pH probe to tell us when the reaction is done

    16. Acid-Base Titrations • We record how much (the volume) of the titrant we put in to make the reaction complete • From there we can figure out the number of moles of titrant we used • We use our knowledge of acids and base reactions to relate that information to the number of moles in our unknown • We use that plus the volume of unknown we put in the flask to calculate the concentration of the unknown sample

    17. Acid-Base TitrationsIndicators • When we titrate, we need a way of determining when the reaction is finished • If we know at which pH the end of the reaction should be… • we can use a chemical that changes colour at that pH… • then we can use that colour change to indicate when our reaction is complete • The point where the indicator changes colour is called the “endpoint” of the reaction • Because it signals that end of the reaction

    18. Acid-Base TitrationsIndicators • In order for an “indicator” to work, we must choose one that changes colour when the equivalence point is reached (the end of our reaction) • Which means we must calculate where that point should be in order to pick the right one! • Why does an indicator change colour? • When there is no more analyte (solution in the flask) to react with, the titrant (solution in the buret) will react with the indicator instead • This reaction causes an indicator to go from an acid/base to its conjugate base/acid • When it does that, it changes colour H(Indicator) + H2O ⇄ (Indicator)- + H3O+ Acid formBase form An indicator is a conjugate weak acid/base pair!

    19. Acid-Base TitrationsIndicators

    20. Acid-Base TitrationspH Probes • A pH probe/meter goes into the analyte solution • It is connected to a computer • As the titrant is added, the pH is measured by the probe and the data is recorded by the computer • It is a more accurate way of determining the end of a reaction

    21. Acid-Base Titrations McGraw-Hill Animation

    22. Acid-Base TitrationsTitration Curves Titration Curves A plot of pH versus volume of titrant added (solution in buret with known concentration) Strong Acid titrated with Strong Base As the volume of base added increases, the pH increases slightly UNTIL… The point where there is no more H+ from HCl left to react with = neutralization Then as more NaOH is added, the pH rises dramatically as [OH-] dominates the solution

    23. Acid-Base TitrationsTitration Curves Titration Curves The “Equivalence Point”-the point at which exactly enough titrant has been added to react with all analyte All titrant added after the equivalence point has no analyte left to react with In this case, with NaOH being our titrant, the pH rises quickly in response to the addition of more and more base Midpoint of the sharp change in pH

    24. Acid-Base TitrationsTitration Curves Because this is a strong acid with a strong base, the pH at the equivalence point is “neutral pH” = 7 This is NOT the case when we mix a weak acid/base with a strong base/acid

    25. Acid-Base Titrations-Titration Curves Strong Acid titrated with Strong Base Weak Base titrated with Strong Acid Weak Acid titrated with Strong Base

    26. Acid-Base TitrationsTitrations and Titration Curves What kind of calculations can we do with a titration if both concentrations are known? We can calculate pH before adding titrant We can calculate pH after a specified volume of titrant added We can calculate the pH at the equivalence point We can calculate pH after the equivalence point We can calculate the amount of moles of titrant or analyte at any point during the titration

    27. Acid-Base TitrationsTitrations and Titration Curves What kind of calculations can we do with a titration curve? We can calculate the concentration of an unknown solution We can calculate the amount of moles of titrant at any point during the titration We can calculate the Ka for a weak acid titrated with a strong base We can calculate the Kb for a weak base titrated with a strong acid

    28. Acid-Base Titrations • When we titrate, there are 3 different scenarios we look at: • A strong acid and a strong base • A strong acid with a weak base • A strong base with a weak acid • For each case, the “titration curve” will have a similar wavy shape but the reaction will end at a different pH value

    29. Strong Acid/Strong Base Titration How would we choose an indicator for this reaction? We would look for one that has its colour change around pH 7 Bromthymol Blue

    30. Strong Acid and Strong Base Titration Calculations • When we titrate a strong acid with a strong base, we end up with a neutral solution: NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) Strong base Strong Acid Neutral Salt Water pH mL of NaOH added

    31. Strong Acid and Strong Base Titration Calculations NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) Strong base Strong Acid Neutral Salt Water • If we have NaOH in the buret and HCl in the flask • as we add NaOH, the OH- reacts with the H+ from HCl to form water-to “neutralize” the solution • With every drop of NaOH added, we end up with less H+. • Which causes our pH to go up (become more basic)

    32. Strong Acid and Strong Base Titration Calculations NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) Strong base Strong Acid Neutral Salt Water Formula Equation: NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) Complete Ionic Equation: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)→ Na+(aq) + Cl-(aq) + H2O(l) Net Ionic Equation: OH-(aq) + H+(aq) → H2O(l) OH-(aq) + H3O+(aq) → 2H2O(l) Tells us what is actually reacting in the beaker

    33. Strong Acid and Strong Base Titration Calculations NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) • Net Ionic Equation: OH-(aq) + H+(aq) → H2O(l) Example Calculation 1: In a titration, 20.00mL if 0.300mol/L HCl(aq) is titrated with 0.300mol/L NaOH(aq). What is the amount of UNREACTED HCl(aq) and the pH of the solution after the following volumes of NaOHare added? • 0.00 mL initial solution pH • 10.0mL • 20.0mL • 25.0mL

    34. Strong Acid and Strong Base Titration Calculations NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) • 0.00mL Initial pH HCl(aq) + H2O(l) ⇄ H3O+ + Cl- Since HCl is a strong acid and ionizes completely, the concentration of H3O+(aq) before NaOH is added will be equal to the initial concentration of HCl [HCl] = [H3O+] pH = -log[H3O+] pH = -log(0.300) pH = 0.5

    35. Strong Acid and Strong Base Titration Calculations NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) • 0.00mL Initial pH HCl(aq) + H2O(l) ⇄ H3O+ + Cl- All HCl is unreacted at this point, so we figure out how many moles that is nHCl= C x V = 0.300 mol/L x 20.0mL = 6.00 x 10-3 moles unreacted Which is equal to the number of moles of H+ and Cl- since HCl completely dissociates into its ions

    36. Strong Acid and Strong Base Titration Calculations • 10.0mL of 0.300mol/L NaOH When NaOH is introduced into the flask, the result is a reaction between the H+ from HCl with the OH- from NaOH OH-(aq) + H+(aq) → H2O(l) Step 1: Calculate how many moles of OH- were introduced into the flask n = C * V n = 0.300M * 10 x 10-3L n = 3.00 x 10-3 mol OH-

    37. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) Step 2: Look at the stoichiometry of the reaction to determine how many moles of H+ will react with the OH- Since it is a 1:1 ratio, 3.00 x 10-3 moles of H+ will react with the OH- from the NaOH Step 3: Subtract from the initial amount of moles of H+ 6.00 x 10-3 – 3.00 x 10-3 = 3.00 x 10-3 moles of H+unreacted

    38. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) Step 4: Determine the concentration of unreacted H+ The new volume of solution is 30.00mL (20.00mL from the HCl and 10.00mL from the NaOH added) The new concentration is then: C = n/V C = 0.100mol/L

    39. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) Step 5: Determine the pH Since it is the unreacted H+ in solution that governs the pH, use its concentration in the pH equation to determine the pH pH = -log[H+] note H+ and H3O+ are the same thing pH = -log(0.100) pH = 1.00

    40. Strong Acid and Strong Base Titration Calculations • 20.0mL of 0.300mol/L NaOH Your turn OH-(aq) + H+(aq) → H2O(l) Step 1: Calculate how many moles of OH- were introduced into the flask n = C * V n = 6.00 x 10-3 mol OH-

    41. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) Step 2: Look at the stoichiometry of the reaction to determine how many moles of H+ will react with the OH- Since it is a 1:1 ratio, 6.00 x 10-3 moles of H+ will react with the OH- from the NaOH Step 3: Subtract from the initial amount of moles of H+ 6.00 x 10-3 – 6.00 x 10-3 = 0 moles of H+unreacted

    42. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) Step 4: Determine the concentration of unreacted H+ The new concentration is then: C = n/V C = 0.00mol/L

    43. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) Step 5: Determine the pH Since there is no unreacted H+ in solution, the only thing left is H2O(l), Na+ and Cl- The equivalence point has been reached and the solution has become neutral pH = 7

    44. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) • 25.0mL of NaOH We know that it takes 20.0mL of NaOH to react with all the HCl in solution. We have 5.0mL of 0.300mol/L NaOH left over that is going to effect the pH nNaOH = nOH- = C x V nOH- = 0.0015 mol OH- COH- = 0.0015 mol / Vtotal COH- = 0.0015 mol/ 55 x 10-3 L COH- = 0.0273 mol/L

    45. Strong Acid and Strong Base Titration Calculations OH-(aq) + H+(aq) → H2O(l) COH- = 0.0273 mol/L pOH = -log[OH-] pOH = -log(0.0273) pOH = 1.564 pH = 14 - pOH pH = 14 - 1.564 pH = 12.4

    46. Strong Acid and Strong Base Titration Calculations Calculations from a Titration Curve NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l) OH-(aq) + H+(aq) → H2O(l) Example 2: If, according to the plot, it took 20mL of 0.100mol/L NaOH to neutralize 20.0mL HCl, what is the molarity of HCl in the flask?

    47. Strong Acid and Strong Base Titration Calculations Calculations from a Titration Curve OH-(aq) + H+(aq) → H2O(l) If it took 20mL of 0.100mol/L NaOH to neutralize the HCl, what is the molarity of HCl in the flask? Step 1: Determine the amount of moles of OH- it took to reach the equivalence point nNaOH = nOH- = C x V = 0.002 mol OH-

    48. Strong Acid and Strong Base Titration Calculations Calculations from a Titration Curve OH-(aq) + H+(aq) → H2O(l) Step 2: Look at the stoichiometry of the reaction to determine the number of moles of H+ that reacted with the OH- 1:1 therefore, Moles H+ = moles OH- = 0.002 mol H+

    49. Calculating Unknown Concentrations with Strong Acid and Strong Base Titrations Calculations from a Titration Curve OH-(aq) + H+(aq) → H2O(l) Step 3: Use stoichiometry to determine the number of moles of HCl that correspond to the number of moles of H+ HCl(aq) → H+(aq) + Cl-(aq)1:1 Moles HCl(aq) = H+(aq)= moles OH-(aq)= 0.002 mol HCl(aq) Step 4: Find Initial HCl(aq) Concentration CHCl = n/V CHCl = 0.002mol/20 x 10-3 L CHCl = 0.100 mol/L

    50. Weak Acid/Strong Base Titration How would we choose an indicator for this reaction? We would look for one that has its colour change around pH 9 Phenolphthalein Now we have an equilibrium to think about!