Presentation on theme: "Topic 1C – Part II Acid Base Equilibria and Ksp"— Presentation transcript:
1Topic 1C – Part II Acid Base Equilibria and Ksp Applying EquilibriumTitrationsBuffersSolubility Product
2TitrationsTitration is an experimental technique used to perform a neutralization reaction (see unit 3).Used to determine the unknown concentration of an acid or base.As a base or acid is added to an acid or base in a titration, there is very little change in pH initiallyWhen the moles of titrant (from the buret) are in the exact stoichiometric proportion with the analyte (in the flask)(100% of the acid or base has been neutralized)This is the equivalence point has been reached.At this point there is a rapid change in pH
3These changes can be summarized using titration curve plots.
5We use an indicator, a chemical that changes colors at various pH’s. Most acid-base titrations involve the addition of one colorless solution to another colorless solutionWe need a method of determining when the equivalence point has been reached.We use an indicator, a chemical that changes colors at various pH’s.Indictors are often weak acids,the ionized and the unionized form have different colors.will change color over a small, given range of pH, for example;
6The equivalence point is located in the middle of the vertical portion of the titration curves. We need to choose an indicator that changes color at a pH value as close to the equivalence point as possible
7This observable color change of the indicator is called the end point, Should correspond to the equivalence point as closely as possible.Using the titration plots and the table on the previous slide, suitable indicators can be chosen;Strong acid - Strong base most indicatorsWeak acid - Strong base phenolphthaleinStrong acid - Weak base methyl orange
8Titration Calculations Millimole (mmol) = 1/1000 molMolarity = mmol/mL = mol/LMakes calculations easier because we will rarely add Liters of solution.Adding a solution of known concentration until the substance being tested is consumed.This is called the equivalence point.Graph of pH vs. mL is a titration curve.
9Strong acid with Strong Base Do the stoichiometry.There is no equilibrium .They both dissociate completely.The titration of 50.0 mL of M HNO3 with M NaOHCalculate the pH after the following ml of NaOH are added:A. no NaOHB. 10ml of NaOHC. 20ml of NaOHD. 50ml of NaOHE. 100ml of NaOHF. 150ml of NaOH
10Weak acid with Strong base There is an equilibrium.Do stoichiometry.Then do equilibrium.Titrate 50.0 mL of 0.10 M HC2H3O2(Ka = 1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at the following points.A. No NaOHB. 10ml of NaOHC. 25ml of NaOHD. 40ml of NaOHE. 50ml of NaOHF. 60ml of NaOH
11Titration Curves 7 pH mL of Base added Strong acid with strong Base Equivalence at pH 77pHmL of Base added
12>7 pH mL of Base added Weak acid with strong Base Equivalence at pH >7>7pHmL of Base added
13Strong base with strong acid Equivalence at pH 77pHmL of Base added
14<7 pH mL of Base added Weak base with strong acid Equivalence at pH <7<7pHmL of Base added
15Summary: Titration of a Weak Acid with a Strong Base ** At 0 ml base find pH with ICE and Ka** Weak acid before equivalence pointStoichiometry firstThen Ka Equation or Henderson-Hasselbach** Weak acid at ½ equivalence pt[H+] = [Ka] or pH=pKa** Weak acid at equivalence point find pH with ICE and Kb** Weak base after equivalence - leftover strong base.
16Indicators Weak acids that change color when they become bases. weak acid written HInWeak baseHIn H+ + In-clear redEquilibrium is controlled by pHEnd point - when the indicator changes color.
17Indicators Since it is an equilibrium the color change is gradual. It is noticeable when the ratio of [In-]/[HIn] or [HIn]/[In-] is 1/10Since the Indicator is a weak acid, it has a Ka.pH the indicator changes at is.pH=pKa +log([In-]/[HIn]) = pKa +log(1/10)pH=pKa - 1 on the way up (adding a base)
18Indicators pH=pKa + log([HIn]/[In-]) = pKa + log(10) pH=pKa+1 on the way down (adding an acid)For strong acid base titrations, the indicator color change is sharp and a wide range of indicators , might be suitable.For titrations of weak acids choose the indicator with a pKa as close to equivalence point as possible.
19The Common Ion EffectWhen the salt with the anion of a weak acid is added to that acid,It reverses the dissociation of the acid.Lowers the percent dissociation of the acid.The same principle applies to salts with the cation of a weak base.The calculations are the same as last chapter.
20HF (aq) H+ (aq) + F- (aq)If NaF is added to the solution[F-] increases and the equilibrium shifts leftThis makes the solution less acidicSoln of NaF + HF less acidic than soln of HF
21Buffered solutions A solution that resists a change in pH. Either a weak acid and its salt or a weak base and its salt.We can make a buffer of any pH by varying the concentrations of these solutions.Same calculations as before.
2215.2 – The pH of a Buffered Solution A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2).Calculate the pH of the solution.
23Adding a strong acid or base Do the stoichiometry first.A strong base will grab protons from the weak acid reducing [HA]0A strong acid will add its proton to the anion of the salt reducing [A-]0Then do the equilibrium problem.
2415.3 – pH Changes in Buffered Solutions Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of the buffered solution in sample exercise 15.2.
25Buffers – How do they work? A buffered solution contains large quantities of a weak acid (HA) and its conjugate base (A-) from the salt.When OH- ions are added they pull off the H+ from the HAOH- + HA A- + H2OOH- cannot accumulate. It reacts with HA to produce A-
26General equation Ka = [H+] [A-] [HA] so [H+] = Ka [HA] [A-] The [H+] depends on the ratio [HA]/[A-]taking the negative log of both sidespH = -log(Ka [HA]/[A-])pH = -log(Ka)-log([HA]/[A-])pH = pKa + log([A-]/[HA])
27This is called the Henderson-Hasselbach equation pH = pKa + log([A-]/[HA])pH = pKa + log(base/acid)
2815.4 – The pH of a Buffered Solution II Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate.15.5 – The pH of a Buffered Solution IIIA buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.
2915.6 – Adding Strong Acid to a Buffered Solution I What would the pH be if 0.10 mol of HCl is added to 1.0 L of the buffered solution from Ex
30Buffering CapacityThe pH of a buffered solution is determined by the ratio [A-]/[HA].As long as this doesn’t change much the pH won’t change much.The more concentrated these two are the more H+ and OH- the solution will be able to absorb.Larger concentrations bigger buffer capacity.
3115.7 – Adding Strong Acid to a Buffered Solution II Calculate the change in pH that occurs when mol of HCl(g) is added to 1.0L of each of the following:5.00 M HAc and 5.00 M NaAc0.050 M HAc and M NaAcKa= 1.8x10-5
32Buffer capacity The best buffers have a ratio [A-]/[HA] = 1 This is most resistant to changeTrue when [A-] = [HA]Make pH = pKa (since log1=0)
3315.8 – Preparing a BufferA chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts):a. chloroacetic acid (Ka = 1.35 x 10-3)b. propanoic acid (Ka = 1.3 x 10-5)c. benzoic acid (Ka = 6.4 x 10-5)d. hypochlorous acid (Ka = 3.5 x 10-8)Caclulate the ratio [HA] / [A-] required for each to yield a pH of Which system will work best?
34Solubility Equilibria (Equilibria between solids and solutions) All dissolving is an equilibrium.If there is not much solid it will all dissolve.As more solid is added the solution will become saturated.Solid dissolvedThe solid will precipitate as fast as it dissolves .Equilibrium
35Consider the following equilibrium CaF2 (s) Ca2+ (aq) + 2F- (aq)Ultimately this reaches equilibriumSolubility Product Constant (Ksp) isKsp = [Ca2+] [F-]2Called the solubility product.
36Solubility is not the same as solubility product Solubility product is an equilibrium constant.it doesn’t change except with temperature.Solubility is an equilibrium position for how much can dissolve.A common ion can change this.
3715.12 - Calculating Ksp from Solubility I Copper (I) Bromide has a measured solubility of 2.0 x 10-4 mol/L at 25oC.Calculate its Ksp value
3815.13 - Calculating Ksp from Solubility II Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x mol/L at 25oC.
3915.14 - Calculating Solubility from Ksp The Ksp value for copper (II) iodate, Cu(IO3)2, is 1.4 x 10-7 at 25oC. Calculate its solubility at 25oC.
40Relative solubilities Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions.The bigger the Ksp of those the more soluble.If they fall apart into different number of pieces you have to do the math.
41Common Ion Effect 15.15 - Solubility and Common Ions If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.Solubility and Common IonsCalculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a M NaF
42pH and solubility OH- can be a common ion. More soluble in acid or less in base.Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.H+ + PO HPO42-Ag3PO4 (s) Ag+ (aq) + PO43- (aq)
43Precipitation and Qualitative Analysis Ion Product, Q =[M+]a[N-]bIf Q>Ksp a precipitate forms.If Q<Ksp No precipitate.If Q = Ksp equilibrium.15.17 – Determining Precipitation ConditionsA solution is prepared by mixing mL of 1.00 x 10-2 M Mg(NO3)2 and mL of1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9).
44Salts as acids and bases Salts are ionic compounds.Salts of the cation of strong bases and the anion of strong acids are neutral.for example:NaCl (salt) – NaOH or HClKNO3 (salt) – KOH or HNO3There is no equilibrium for strong acids and bases.Na+, K+ and Cl-, NO3- have no affinity for OH- ions or H+ ions.
45Salts that Produce a Basic Soln If the anion of a salt is the conjugate base of a weak acid – this produces a basic solution.In an aqueous solution of NaF (Salt)F- (anion) / HF (weak acid)The major species are Na+, F-, and H2OF- + H2O HF + OH-Kb = [HF][OH-][F- ]
46Ka tells us Kb Ka x Kb = KW = 1.0 x 10-14 The anion of a weak acid is a strong base.Calculate the pH of a solution of 0.30 M NaF solution.(Ka of HF is 7.2 x 10-4)(Ex )The F- ion competes with OH- for the H+
47Acidic saltsA salt with the conj. acid of a weak base and the anion of a strong acid will be acidic.NH4Cl (salt)NH4+ (conj. acid of weak base – NH3)Cl- (anion of a strong acid - HCl)Calculate the pH of a solution of 0.10 M NH4Cl (Kb of NH3 1.8 x 10-5) (Ex )Other acidic salts are those of highly charged metal ions.
48Anion of weak acid, cation of weak base If the salts in solution produce both cations and anions with acidic and basic properties.Ka > Kb acidicKa < Kb basicKa = Kb NeutralExample: NH4C2H3O2 or NH4CN
49The Lewis Acid-Base Model Definition of AcidDefinition of BaseArrheniusH+ producerOH- producerBronsted-LowryH+ donorH+ acceptorLewisElectron-pair acceptorElectron-pair donor
50Lewis AcidHas an empty orbital that can accept an electron pairLewis BaseHas a lone pair of electrons that can be shared