Common Ion Effect

Other Aspects of EquilibriaBuffers Common Ion Effect (explained by Le Chatelier) When a salt with an anion of a weak acid is added to that weak acid. 1.0 HF reacts with 1.0 M NaF  More F- ions will shift equilibrium to left.  Reverses the dissociation of acid and lowers percent dissociation. Same thing applies to salts with cations of a weak base added to that weak base.Calculations are the same as last chapter! - -  Initial [A ] is not zero because the solution also contains the salt of A ).Example Calculate [H+] and % dissociation of HF in a solution containing 1.0 M HF (Ka =7.2 x 10- 4) and 1.0 M NaF.  Compare with results of just [HF] = 1.0 where [H+] = 2.7 x 10-2 or 2.7%Buffer Solutions  One that resists change in the pH when a small amount of strong acid or strong base is added to it. Usually made from a weak acid and one of its salts. Or can be made from weak base and one of its salts Examples: HC2H3O2 and NaC2H3O2 OR NH3 and NH4Cl+ -1  Major species Na , C2H3O2 , HC2H3O2 and H2O -1 -1 + C2H3O2 : Absorbs excess acid: C2H3O2 + H3O  HC2H3O2 + H2O   - - HC2H3O2 : Absorbs excess base: HC2H3O2 + OH  C2H3O2 + H2O    “Mopping Up”  pH remain unchangedExample: Calculate the pH of a Solution that is .50M HC2H3O2 and .25M NaC2H3O2 (Ka= 1.8 x 10-5).Adding a Strong Acid  decrease pH or Strong base  increase pH1. Assume the reaction goes to completion(all acid or base is used up) 2. Do stoichiometry first- use neutralization reaction + 3. Strong base will grab H from weak acid, reducing initial concentration. [HA]o - 4. Strong acid will add H+ to anion of the salt reducing initial concentration. [A ]o 5. Do equilibrium problem – use Ka or Kb and new concentrationsExample: R - - E HA + OH  A + H2O } rxn is complete M weak acid strong base E M B A- + H+  HA } reaction is complete E weak base strong acid RExample Calculate the pH that occurs when .010 mol NaOH is added to 1.0L of .50M HC2H3O2 / . 50M NaC2H3O2 buffer. -5 (Ka [HC2H3O2 ]= 1.8 x 10 )Another Method: Henderson Hasselbach+ - +  General equation: Ka = [H ][A ] so [H ] = Ka [HA] [HA] [A-] +  [H ] depends on ratio Ka and [HA] if you take the (–) log of both sides [A-] if working with an acid if working with a base- +  pH = pKa + log [A ] / [HA] pOH = pKb + log [HB ] / [B] pH = pKa + log (base / acid) pOH = pKb + log (acid / base) Example: -4 Calculate the pH of a solution containing .75 M lactic acid , HC3H5O3 (Ka = 1.4 x 10 ) and .25 M sodium lactate, Na C3H5O3Another example: -5 A buffered solution contains .25 M NH3(Kb = 1.8 x 10 ) and .40 M NH4Cl. Calculate the pH.Buffer Capacity  the amount of acid or base the buffer can neutralize before pH changes to an appreciable amount  the pH of a buffered solution is determined by the ratio [A-] / [HA] ; as long as ratio is large compared to amount of OH- added, the ratio will not change much  as long as this ratio does not change much, the pH does not change much  the more concentrated these two are, the more H+ and OH- the solution will be able to absorb  large concentrations- big buffer capacityExample: Calculate the change in pH that occurs when .010 mol HCl is added to 1.0 L of each of -5 the following: (Ka= 1.8 x 10 ) a) 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 b) .050 M HC2H3O2 and .050 M NaC2H3O2 Best Buffers have a ratio of [A-] / [HA] = 1  This is true when [A-] = [HA] This makes the pH = pKa since log 1 = 0 pKa should be as close to desired pH as possible- Example: pH = pKa + log [A ] / [HA] -4 pH of 4.0 is wanted so 4.00 = pKa and thus Ka = 1.0 x 10Solubility Equilibria  Dissolving a solute into a solvent will eventually cause saturation  Solid ↔ Dissolved solid  This a type of physical equilibrium  solids precipitate as fast as they dissolveIonic Compound+ - Ma NMb (s)  aM (aq) + bNM (aq)+2 - CaCl2 Ca (aq) + 2 Cl (aq)+2 1 - 2 So the equilibrium expression is Ksp = [Ca ] [Cl ] Remember so pure solids or liquids are placed in the equilibrium expression Ksp = Solubility product (equilibrium constant) - only changes with temperature changeHow to calculate Ksp? Example: -4 Calculate Ksp of copper(I) bromide which has a measured solubility of 2.0 x 10 M @25°C.-154 Calculate Ksp for bismuth sulfide (Bi2S3) which has a solubility of 1.0 x 10 M @25°C.WATCH OUT! Solubility is not the same as the solubility product constant. It is the equilibrium position for how much can dissolve!How to calculate solubility?Example: -7 Ksp for copper iodate Cu(IO3)2 is 1.4 x 10 @ 25 C. Calculate its solubility? Ksp will only allow you to compare the solubility of solids that fall apart into the same number of ions (NaCl, KF…) The larger the Ksp, the more soluble!  If they fall into different number of ions- must do the math!Example: Put in order from greatest to least solubility! -16 AgI Ksp = 1.5 x 10 -12 CuI Ksp = 5.0 x 10 -5 CaSO4 Ksp = 6.1 x 10Common Ion Effect  Dissolving a solid into a solution will either the cation or anion already present will cause less to dissolve – Le Chatelier’s PrincipleExample -11 Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10 ) in a .025 M NaF solution. Precipitation Predictions  Use ion product, Q = [M+]a [NM-]b  use [initial] If Q> Ksp precipitation If Q < Ksp no precipitation If Q = Ksp at equilibriumExample -3 A solution is prepared by adding 750.0 ml of 4.00 x 10 M Ce (NO3)3 to 300.0 ml of 2.00 -2 x 10 M KIO3. -10 Will Ce (IO3)3 with a Ksp = 1.9 x 10 precipitate from solution? If you want to calculate equilibrium concentrations in solution after precipitation occurs 1) must determine Q to see if its takes place 2) reverse dissociation equation – do stoichiometry- and assume it goes to completion 3) Adjust back to equilibriumExample: -2 A solution is prepared by mixing 150.0 ml of 1.00 x 10 M Mg(NO3)2 and 250.00 ml of 1.00 x 10-1 M NaF. +2 -1 Calculate the concentration of Mg and F at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9) Titrations  Technique used to determine the amount of acid or base in solution  Add known concentration of titrant( from buret) until substance being tested is consumed. (Equivalence point)  Equivalence point is where stoichiometrically equivalent amounts have been added  Equivalence point is often determined by color change of an indicator (end point)  End Point (color change) is different than equivalence point( equal amounts have been added)  Titration curve- graph of pH vs. ml added(titrant)  PH meters monitor the acid base reaction.  To make calculations easier, usually work with millimoles 1 mmol = 1/1000 th of a mole Molarity = mol / L = mmol/mlStrong Acid – Strong Base Titrations  A strong acid titrated with a strong base + -  Net ionic: H + OH  H2O  Curve can be thought of as 4 regions: (See Titration Curve 1 on handout) 1) Before any base has been added-pH depends on only strong acid 2) No buffer region 3) Endpoint (equivalence point) is 7.0 always 4) After equivalence point- dependent on left over strong baseCase Study: Titration of 100. ml of .100 M HCl with a .100 M NaOH No NaOH added: HCl  H+ + Cl- 20.0 ml of NaOH added: HCl + NaOH  H2O + NaCl**Because volume has been added in the titration process you must refigure the molarity** 100.0 ml of NaOH added** Equivalence Point is reached**  110.0 ml of NaOH addedWeak Acid/Strong Base Titrations Curve can be thought of 4 regions: (See Titration Curve 2 on handout!) 1. ore any base is added: pH depends on weak acid 2. After base is added- before equivalence point-pH depends on weak acid/salt buffer 3. At equivalence point- hydrolysis of anion of weak acid: pH depends on anion 4. After equivalence point: excess strong base; pH depends on strong base2 steps: 1. stoich problem – assume reaction of [OH-] with weak base runs till completion; acid concentration remaining and conjugate base concentration remaining 2. Equilibrium problem- use weak acid- Case Study: Titration of 50.0 ml of .10 M HC2H3O2 (Ka = 1.8 x 10 5) with .10 M NaOH + - No NaOH added: HC2H3O2 H + C2H3O2 (calculate pH of weak acid) 10.0 ml of NaOH added: (Buffer mixture of acetate and weak acid)- - Stoich calc. OH + HC2H3O2 C2H3O2 +H2O+ - Equilibrium calc. HC2H3O2 H + C2H3O2 50.0 ml of NaOH added ** Equivalence Point is reached** -pH will always be greater than 7 60.0 ml of NaOH addedWeak Bases/Strong Acid Titrations  similar to a weak acid-strong base except invertedWeak Acids/Weak Bases Titrations  change in pH is too gradual near equivalence point for indicators to be usedIndicators  dyes that are weak acids  used to identify equivalence point  changes color at endpoint of a titration + - - +  HIN  H + In Ka = [In ] [H ] Acid Conjugate base [H In] 1 color different colorMust choose indicator with a Ka near that of an acid being titrated and one that changes color over the pH range of the sharp vertical step on each graph:  Strong Acid/Strong Base: most indicators  Strong Acid/Weak Base: methyl orange/methyl red  Weak Acid/Strong Base: phenolphthalien

Common Ion Effect

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