Transcript Slide 1
Chapter 12 Aqueous Equilibrium Common Ion Effect Learning Outcomes and Prereqs Pre-reqs: Identify how changing concentrations of reactants or products shift an equilibrium (Le Chatelierβs Principle) Learning Outcomes Identify the effect of adding a conjugate base of an acid on pH and ionization. Identify the effect of adding a conjugate acid of a base on pH and ionization. Reminder Le Chatelierβs Principle If you apply stress to a system it shifts to relieve the stress. Reaction shifts left or right to relieve the stress. Analogy- The common ion effect If you have a solution of an acid and its conjugate base the ionization is lessened by the presence of the ion. Same is true for a base and its conjugate acid Conjugate acid/base added through itβs salt. Soβ¦.what does that mean? Examples? The common ion effect Adding a conjugate lessons the effect of ionization. Examples? Formic acid and sodium formate H Ammonia and Ammonium Chloride Why do we care? If ionization is decreased how does this affect the OH- and H+ concentrations? H Question: If HCOO- is added, what happens to the concentration of H+? What does that do to pH? [H+]: It decreases pH: It increases But does it really matter? An example to illustrate: Determine the pH of a 0.20M NH3 solution. Then determine the pH of a solution that is 0.20M NH3 and also 0.30M NH4Cl Review If you add the conjugate base of an acid: the ionization will be lessened This means less [H+] so a higher pH If you add the conjugate acid of a base: The ionization will be lessened This means less [OH-] so a lower pH. Henderson Hasselbalch Equation Learning outcomes Derive the Henderson Hasselbalch Equation Use the Henderson Hasselbalch equation to solve a problem. Identify when we are allowed to use this, and when we have to use the previous methods of solving acid/base problems. Henderson-Hasslebalch: Creating the generic equation Henderson-Hasslebalch: Problem solving tips Use this for buffer problems only. Be sure you are using the Ka NOT the Kb As with all equations, get good at manipulating them. Only use if weak acid approximation is valid Example: Using the HendersonHasslebalch equation A solution is made that is 0.20M CH3COOH and 0.30M CH3COONa are mixed. Find the pH. 0.30 ππ» = 4.75 + πππ 0.20 ππ» = 4.93 Review The Henderson Hasselbalch equation comes from solving a generic acid problem where both an acid and its conjugate base are present. You can use this equation to solve problems whenever the 5% approximation is valid (note: you donβt have to show me, but you do need to check in case it doesnβt work) Introduction to Buffers Motivation and Learning Outcomes Identify why our body needs buffers, and what happens if something causes that system to fail. Define buffers. Identify how buffers work to keep pH changes small Identify if two components form a buffer Real life importance check: pH of blood=7.4 Acidosis <7.35 Some products of your metabolism: Ammonia Urea (a base) Uric acid CO2 (acidic when reacting with water) Alkalosis> 7.50 Urea Alkalosis and Acidosis Metabolic Alkalosis and Acidosis Respiratory Hypoventilation increased CO2 acidosis Hyperventilation decreased CO2 alkalosis Real life importance check: pH of blood=7.4 Acidosis <7.35 Alkalosis> 7.50 Products of your metabolism: Ammonia Urea (a base) Uric acid CO2 (acidic when reacting with water) So how can your body control your pH so well? Urea Buffers Acid and its conjugate base, or a base and its conjugate acid Essentially the systems weβve talked about in the common ion effect questions It works by converting a strong acid into a weak acid, or a strong base into a weak base. A strong base canβt exist in solution with a weak acid A strong acid canβt exist in solution with a weak base Activity Part 1: (Iβll need some volunteers) Buffers Acid and its conjugate base, or a base and its conjugate acid Basically the systems weβve talked about in the common ion effect questions It works by converting a strong acid into a weak acid, or a strong base into a weak base. A strong base canβt exist in solution with a weak acid A strong acid canβt exist in solution with a weak base Activity Part 2: (Iβll need some volunteers again) Acids react with bases Bases react with acids Real life importance check: pH of blood=7.4 Alkalosis> 4.5 Acidosis <7.35 Products of your metabolism: Ammonia Urea (a base) Uric acid CO2 (acidic when reacting with water) Urea So how can your body control your pH so well? It uses buffers!!!! Buffers in the Body: We discussed how CO2 can become acid in the body This equilibrium is as follows, if both are present itβs a buffer Buffers in the Body: How do we regulate CO2 levels? Breathing levels What happens to the CO2 if we breath shallowly? Respiratory acidosis Why is this bad? Many reasons: here is one: Hemoglobin in red blood cells binds either O2 or H+ π»ππ» + + π2 β π»ππ2 + π» + Hypoventilation increased CO2 acidosis Hyperventilation decreased CO2 alkalosis Which of the following are buffers: KNO2/HNO2 yes KCl/HCl no KHSO4/ H2SO4 no Na2HPO4/NaH2PO4 Yes KCN/HCN Yes Na2SO4/NaHSO4 Yes NH3/NH4NO3 Yes Review Buffers prevent large changes in pH caused by the addition of acids or bases. This is needed in many systems. Our blood buffer system uses buffers to keep the pH in a very narrow range. Buffers work by having both an acid and its conjugate base, or a base and its conjugate acid present. Therefore if an acid is added, it reacts with the base If a base is added, it reacts with the acid This keeps the pH change small. Buffer Calculations. Learning outcomes Use the Henderson-Hasslebalch equation to solve for the pH of buffers. Use the Henderson-Hasslebalch equation to solve for the pH of buffers after adding an acid or base. Test your knowledge of buffers by answering a couple of conceptual questions. Buffers Calculate the pH of the solution that results from mixing 0.200L of 0.050M (CH3)2NH with 0.320L of 0.040M (CH3)2NH2Cl. (Kb=5.88x10-4) You try: Calculate the pH of the solution that results from mixing 72.0mL of 0.015M (CH3)2NH with 20mL of 0.200M (CH3)2NH2Cl The pH of blood plasma is 7.40, Assuming the principal buffer system is HCO3- and H2CO3 find the ratio of HCO3- and H2CO3. (Ka=4.2x10-7) Adding acids and bases Acids react with bases Bases react with acids Buffers when adding a base or acid It works by converting a strong acid into a weak acid, or a strong base into a weak base. A strong base canβt exist in solution with a weak acid A strong acid canβt exist in solution with a weak base Notice NOT equilibrium arrow!!! Use βnewβ concentration of acid or base to do equilibrium calculations in the same manner as before. Steps for buffer problem with addition of acid or base. Step 1: Decide if it is a buffer Step 2: Write reaction of buffer with strong acid or base Step 3: (if needed) do ICF chart (my take on ICE, its not an equilibrium its βfinalβ) to find amount of weak acid or base Suggested, do this in mols Step 4: fill into the HH equation if assumptions are valid (Ka/Kb is low, concentrations high) or use ICE chart and Ka or Kb ectβ¦as we did with other equilibrium problems. If using HH equation use moles If using Ka or Kb equation use molarity Buffers when adding a base or acid Examples: Calculate the pH of 1.00L of the buffer 1.00M CH3COONa and 1.00M CH3COOH. Ka=4.2x10-7 A) before and after the addition of * B) 0.080 mol NaOH or C) 0.120 mol HCl *assume no change in volume ***Note: Youβll notice your book always uses an ICE chart, filling into K, solving for X, finding pH protocol, Iβm not sure why they use this after having taught you the easier way to do it. Choose whichever you like most. Example Questions: We said the normal ratio of HCO3- :H2CO3 is 20:1, Is your blood better at buffering added acid or added base? Review For most cases you can use the Henderson-Hasslebalch equation for solving buffer problems. (When the 5% rule is valid, you need to test this for buffers, but you donβt need to show me, otherwise use an ice chart). If you add an acid or a base to a buffer, first react the species away ( usually using moles) then solve a HH equation. Chemistry and life segment The biology here will not be tested (however you better know the difference between an alcohol and a carboxylic acid since we talked about that earlier in the class). Little side note while we are on biological pH topics: Hangovers: Dispelling another myth βAcetic acid causes acidosis and this is why you get a hangoverβ Hangovers: ο ο Metabolic pathwayAcetaldehyde can cause migraine strength headaches and vomiting. Acetic acid is harmless Alcohol decreases the Anti-diuretic hormone (adh) Increases urine output, Increases dehydration Depletes electrolytes leading to metabolic acidosis! Titration Introduction Learning outcomes Describe the experimental set up of a titration. Identify reasons for doing a titration. Describe the basics of what is occurring during a titration. Which reactions are occurring in the beaker. Using a titration curve, determine what is being titrated (acid or base) and what the titrant is (acid or base). Experimental Set up Burette= known: Titrant Unknown: In Flask Known: In βburetteβ (tall thin volumetric glassware) Typically βstrongβ For acid/base titration: unknown can be acid or base, but burette must have the opposite. Flask/beaker=unknown: Analyte Why? We know what is in the burette (titrant) We know how much of the known we add We can find out the concentration of the unknown (analyte)! Acidic Analyte © Atkins and Jone Basic Analyte (c)Atkins and Jone Titrations: Exercise http://users.wfu.edu/ylwong/chem/titrationsimulator/ You should play with this a bit on your own as well. Review Titrations are used to determine the concentration of an unknown acid or a base. The unknown (analyte) goes in a flask beneath a burette which contains a species of known concentration (titrant). You can identify whether the analyte is an acid or base by measuring the pH (on a titration curve, by seeing the initial pH). For acid base titrations there will be both an acid and a base Titrant=acid, analyte= base OR Titrant=base, analyte=acid Strong Acid/Strong Base Titrations Learning Outcomes Calculate the pH of a strong acid or base titration at various points: Before the titration has started Before the equivalence point At the equivalence point (spoiler, its always 7 for strong/strong) After the equivalence point Use the amount of titrant added to reach equivalence to calculate the concentration, mol, or grams of analyte General Problem Solving Techniques Neutralization Reactions: React H+ and OH- to completion π» + (ππ) + ππ» β (ππ) β π»2 π(π) Calculate moles of H+ and OH-, and find excess concentration. Use pH= -log[H+] or pH=14+log[OH-] Dance Chemistry Explanation http://cast.es.uci.edu/tltc/Temp/Dance%20Chemistr y/Acid%20Base%20Titration.mp4 Example: Strong/Strong Work style #1 Example: If you titrate 100 mL of 1M HCl and 100 mL of 2M NaOH, what is the pH? 0.100 mol 0.200 mol -0.100 mol -0.100 mol 0 mol 0.1 mol Example: Strong/Strong Example: If you titrate 100 mL of 1M HCl and 100 mL of 2M NaOH, what is the pH? Work style #2 OH-= 2mol/L*0.100L=0.2mol H+= 1mol/L*0.100L=0.1mol Which one goes to 0? H+ How much of the excess is left over? 0.1 mol OHHow do we find pH? 14+log[OH-]= 14+log(0.100mol/0.200L)=13.7 Equivalence True for all titrations: At equivalence: mol of titrant added=mol of analyte initially Example: Using volume to equivalence to find information about analyte A 0.120 g Sample of a base is added to 100. mL of water, it is titrated with 3.00mL of 1.00M HCl, what is the molecular mass of the base? (ππππ πππ π)ππππ‘πππ = (ππππ ππππ)πππππ πππ ππππ ππππ = 1.00 β 0.003πΏ = 0.00300 πππ π ππππ ππππ = ππππ πππ π = 0.00300 πππ πππππ 0.120π π πππππ πππ π = = = 40.0 πππ πππ 0.00300πππ Review Calculating the pH of a strong acid or base titration simply requires finding how much H+ or OH- is in excess and solve as we did our βmixingβ problems. Because initial mols of analyte=added mols of titrant you can determine the mols of starting analyte using what you know about how much titrant was added Use this to solve for concentration, molecular mass ectβ¦.using all previous stoichiometry methods. Weak/Weak Titrations Learning outcomes Describe the basics of what is occurring during a titration. Which reactions are occurring in the beaker Which equilibria are present as the titration is progressing. Using a titration curve, determine what is being titrated (weak/strong and acid/base) and what the titrant is (weak/strong and acid/base) Determine the pH at any point during a titration. Determine information about the analyte using what you are given about the titrant. Activity Part 3: Titrations Types of Titrations Buret Holds βTitrantβ Known concentration Strong Acid/Strong Base Strong Base/Strong Acid Weak Acid/Strong Base Weak Base/Strong Acid Weak Acid/Weak Base Erlenmyer Holds βanalyteβ Unknown concentration Weak Base/Weak Acid Example Problem: Weβll do this as we discuss how to do the problem so you can see it done: A 10.0 mL solution of 0.300M NH3 is titrated with a 0.200M HCl solution. Calculate the pH after the following additions of the HCl solution 0.0mL, 10.0 mL, 15.0mL, 30.0mL First establish what type of titration you are dealing with. Step One: Establish what titration you are dealing with Lets walk through them all together: figures © Atkins and Jones Titrations Very complicated problems, be systematic! Two similar approaches that work well to these problems are detailed in the following slides. Pick youβre a favorite way, you really donβt need to know both. In all cases you must know what type of solution you have and how much of each species. We will do one problem each way. βweakβ Titrations: Approach #1 (Typical approach taught, but honestly students tend to not like this way) Step One: Establish what titration you are dealing with, 1B) if it helps draw a sketch of the graph Step Two: Find the Equivalence Point 2B) if it helps label this on the graph axis Step Three: Based on the question decide what area of the titration you are in and how the problem is treated i.e. weak acid only, weak base only, buffer, strong acid only, strong base only. Step 4: Treat problem just as we did in simpler previous problems, just because itβs a titration doesnβt make it different. Step Two: Find the Equivalence Point (c) Atkins and Jones A 10.0 mL solution of 0.300M NH3 is titrated with a 0.200M HCl solution. Calculate the pH after the following additions of the HCl solution 0.0mL, 10.0 mL, 15.0mL, 30.0mL We do this because it tells is which portion of the graph we are in at any given point. Often you will be asked for it. It is an important point. Finding the Equivalence Point Unnecessary if using approach #1 but youβll often be specifically asked for the equivalence point What is the equivalence point? If you start with acid- its when the moles of base to the moles of initial acid If you start with base- its when the moles of acid to the moles of initial added is equal added is equal base Initial Molesacid=Final molesbase OR Initial Molesbase=Final molesacid Find using MaVa=MbVb (concentration must be in mol/volume) Step Two: Find the Equivalence Point (c) Atkins and Jones A 10.0 mL solution of 0.300M NH3 is titrated with a 0.200M HCl solution. Calculate the pH after the following additions of the HCl solution 0.0mL, 10.0 mL, 15.0mL, 30.0mL We do this because it tells is which portion of the graph we are in at any given point. Often you will be asked for it. It is an important point. ππ ππ = ππ ππ (0.200π)ππ = 0.300π (10.0ππΏ) ππ = 15.0ππΏ Based on the question decide what area of the titration you are in and how the problem is treated Not required if using approach 1: however, its extremely useful to be able to be able to recognize the regions and you will be asked about the regions and the regions based on given equivalence points. Beginning: weak base Buffering region: buffer Equivalence point: weak acid 15mL All base is used up and doesnβt play a role, now it is strong acid problem Titrations Step 4 (both approaches): Treat problem just as we did in simpler problems We will do this on the document camera! Step 4: Treat problem just as we did in simpler problems, 0.0 mL HCl added Only weak base is present, treat as weak base problem Since kb will be used, youβll need to use concentration 15mL Step 4: Treat problem just as we did in simpler problems, 10.0 mL HCl added In buffering region: Use HH equation: so use mols (you can use concentration but it is more complicated due to changing volumes) 15mL Step 4: Treat problem just as we did in simpler problems, 15.0 mL HCl added At equivalence point, Same amount of strong acid added as weak base to start This means all the weak base has been converted to a weak acid Treat as weak acid, use ka be sure to adjust concentration (using m1v1=m2v2) use concentration since you are using the k 15mL Step 4: Treat problem just as we did in simpler problems, 30.0 mL HCl added All of the weak base was changed into weak acid, and then more strong acid was added. The weak acid doesnβt have much of an effect as the strong acid Treat as strong acid problem Need to figure out concentration though, some of it was used to convert the weak base to weak acid, subtract that Find volume of HCl past equivalence, use m1v1=m2v2 to find new concentration and fill in to pH=-log(H+) 15mL βweakβ Titrations: Approach #2: (This is not the traditional method taught, however most students appear to prefer it) Step One: Establish what titration you are dealing with, Step Two: Write and equation that describes how the titrant (in buret) and analyte (in beaker) react. Step Three: Make ICF chart and react system to completion (till one of reactants equals zero). Step Four: Look at what is left over. What type of solution do you have? weak acid only, weak base only, buffer, strong acid only, strong base only. Step 4: Treat problem just as we did in simpler problems, just because itβs a titration doesnβt make it different. Example Problem: Repeating the problem as you would with approach number 2: A 10.0 mL solution of 0.300M NH3 is titrated with a 0.200M HCl solution. Calculate the pH after the following additions of the HCl solution 0.0mL, 10.0 mL, 15.0mL, 30.0mL Note: A weak acid titrated with a strong base is very similar to before: Youβll do one of these in discussion. Review Titrations of a weak analyte require an understanding of what part of the titration curve you are in before solving for a pH. In both methods for solving you are required to find out what type of solution you are dealing with after equilibrium has been reached. In method 1 this is done by comparing the volume to equivalence In method 2 this is done by reacting the species to completion using stoichiometry. The equivalence point can be used to solve for initial moles of titrant exactly like we did in strong acid and base calculations (we did not do an example for weak acids/bases, but itβs the same process) Titration Example Questions Learning outcomes Describe the basics of what is occurring during a titration and answer questions based on this knowledge: For example: Which reactions are occurring in the beaker Which equilibria are present as the titration is progressing. Answer questions about how initial conditions change the equivalence point, and how percent purity of the analyte change the equivalence point Identify different types of titrations and describe why they are different. Estimate (< or > 7) the pH of the equivalence point of a titration Identify the pKa by looking at a titration curve. Question 1: If you are doing a strong acid/strong base or strong base/strong acid titration is there a buffer area? Why? No!!! If a strong acid and base are combined they neutralize each other (turning into water) rather than create a weak acid or base which can then dissociate. Question 2 If a generic weak acid βHAβ is titrated with a strong base NaOH, what is the major species at each labeled point? A) HA B) HA and AC) AD) A- +OH- Question 3 What is the Ka of the acid? Estimate based on the graph. pH=pKa at half equivalence About 3.5 ish. Candy demo Question 4 The analyte has 2 grams of base in 100mL of water. How does the volume needed to reach equivalence change if the 2 grams are instead dissolved in 50mL of water? No change What if it is dissolved in 100mL of water? No change What if 1 gram is used instead? It halves What if 4 grams are used instead? It doubles Will the pH at equivalence be higher or lower than 7? more acidic since all base is converted to its conjugate acid. pH<7 Polyprotic acid titrations Polyprotic acids: Mostly conceptual since the math is tedious Learning Outcomes Identify the major components present at each point in a poly protic acid titration. Identify the pKas of the acid AND be able to associate each Ka with its appropriate equation. Use volume to an equivalence point to determine the volume to other equivalence points, and initial information about the titration. Note: We will not be doing any of the calculating pH for this, while technically itβs the exact same as normal weak acid titrations it gets very tedious. If you are going into analytical chemistry it may be worth spending some time learning how to do this, see me in office hours for explanations. Poly Protic Acids A2- HA- HA-/A2- H2A H2A/HA- Questions A2- HA- What is the pKa associated with the removal of the first H? (pKa1) 1.23 What is the Ka associated with the removal of the second H? (pKa2) HA-/A2H2A/HA- H2A 4.19 Questions A2- HA- HA-/A2H2A/HA- H2A If the volume of titrant to reach point D is 50.0 mL, what is the volume required to reach point B? Point A? B: 25.0 mL A: 12.5 mL If it takes 0.1 moles to reach the first equivalence point, how many moles of analyte were there? How many moles (total added) will it take to reach the second equivalence point? ** Analyte: 0.1 mol Second: 0.2 mol **Note: Iβm doing this in mols to speed up the discussion, of course I could also give you this in volume and concentration. Review Polyprotic acids titrations have two equivalence points, and two half equivalence points. This is because they have different Ka values for each hydrogen. The moles required to reach the first equivalence point is equal to the moles of initial analyte. The moles required to reach the second equivalence point is equal to 2x the moles of initial analyte. Indicators Learning Outcomes Define an indicator. Choose an indicator for a titration when given a chart of possible indicators, or the pKa of possible indicators. Choose it based either on the equivalence point of the titration or based on the complete titration curve. Define an end point, and explain the difference between an equivalence point and an end point. Indicators Red Cabbage Indicator Red Cabbage Indicator Purple Cabbage Indicator https://www.youtube.com/watch?v=K8opH7qprX8 Good: Universal indicator (like pH paper) Bad: No sharp contrast like in other indicators such as phenolthalein End point vs equivalence point Equivalence point. When moles of titrant added = moles of analyte initially End point When the color of the indicator changes color We want these to be in the same place!!!! A βgoodβ indicator Changes color at the equivalence point. Aka- end point and equivalence point is the same place. Which of these is a good indicator? Methyl Red Phenolphthalein Could you use Thymol blue? Not perfectly but it could workβ¦. Using pKa. Indicators change color when pH=pKa. We want indicators to change color at the equivalence point. Therefore pKa of indicatorβ equivalence point of analyte. Example: Picking Indicator Based on pKa. pKa of indicatorβ equivalence point of analyte. Your stock room has available Methyl Orange (pka=3.47) , and phenolphthalein (pka=9.3). For the titration shown, which is best? phenolphthalein Review Your indicator should change color at the equivalence point. Aka: the end point and the equivalence point should be similar. To make this happen, the pKa of your indicator should be approximately equal to the pH at the endpoint of your titration. Due to the big jump in pH once the buffering region is past, this normally gives a wide range of possibilities. Introduction to Solubility Equilibrium Just another K Learning outcomes Define solubility equilibrium. Relate solubility equilibrium to other forms of equilibrium we have discussed. Define Ksp and write equations for Ks given a slightly soluble compound. Define Molar Solubility Solubility Equilibrium Not all substances are soluble Many substances are only slightly soluble Ksp the new version of K, same as all the other Products over reactants Now reactants are solids not aqueous though, are they included? As with other it is a dynamic equilibrium Reminder on Dynamic Equilibrium Constantly changing Constant precipitation, and dissolving Total amount dissolved stays constant. (after eq is reached) Why doesnβt the solid matter in the K equation here? Solubility Product Examples Molar SolubilityThe amount of a compound that is soluble. Mols/L Calculated from Ksp Usually designated βsβ, I use x as weβve been doing with equilibrium constants. Why? A handwritten s looks very much like a handwritten 5. Problem solving with solubility equilibrium. Learning outcomes Luckily you have done most of this already with other equilibria Given the Ksp we can find the concentrations of ions Same as Ka, Kb, Keq ectβ¦, only easier since there is no denominator. Given the concentration of an ion we can find the ksp Same as Ka, Kb, Keq ectβ¦, only easier since there is no denominator Use Molar solubility to solve for Ksp Use Molar solubility to solve for concentrations of ions. On your own You have done most of this already with other equilibria Given the Ksp we can find the concentrations of ions Same as Ka, Kb, Keq ectβ¦, only easier since there is no denominator. Given the concentration of an ion we can find the ksp Same as Ka, Kb, Keq ectβ¦, only easier since there is no denominator This is absolutely identical to every other equilibrium we have done, only easier. I will let you do these problems on your own and focus on molar solubility. One note: most people choose not to make an ice chart for these, that is totally acceptable. Molar Solubility: Examples Basically the same as previous equilibrium problems but now using the definition of molar solubility (The amount of a compound that is soluble in mols/L). The molar solubility of MnCO3 is 4.2x10-6M What is the Ksp? If Aluminum Sulfate Ksp is 9.83x10-10 what is the molar solubility? Hint: Most wrong answers on Ksp questions are because its easy to forget to include the coefficients inside the brackets, and easy to forget to exponentiate them given our time spent with acids and bases Molar Solubility and Common Ion Effect Examples How many grams of CaCO3 will dissolve in 3.0x10-3 mL of 0.050 M Ca(NO3)2 . The Ksp is 8.7x10-9. Calculate the molar solubility of AgCl in a 1.00L solution containing 10.0g of dissolved CaCl2 Ksp=1.6x10-10 Review Ksp, like other equilibrium constants are products over reactants (with exponents equal to the coefficients). Because the reactants are solid, and are not included this makes calculations much easier (seldom are there quadratics). Molar solubility, being the amount that a solid can dissolve will allow you to calculate the amount of ions dissolved (using stoichiometric coefficients), and therefore the Ksp. Of course this means that using Ksp, you can solve for the ions dissolved (as done in previous sections) and use that to solve for molar solubility. Common Ion Effect Learning Outcomes Determine if mixing two solutions will cause a precipitate to form. Identify precipitate and spectator ions Using previous problem solving techniques determine the concentration of all ions in solutions Common Ion Effect and Solubility If two solutions which are both soluble are mixed, but any combination of those ions form an insoluble compound a precipitate may form. Lets look at this happening! Demo: CuCl2 and NaOH soluble soluble soluble insoluble Mixing Solutions! We can see if you mix two salt solutions if a precipitate will form. This is like using Q to determine which way the reaction will go. You can have two different salt solutions that are perfectly soluble, but when mixed form a precipitate See the picture for how this happens You can calculate if this will happen Q<K no precipitate Q=K saturated solution Q>K precipitate Demo: CuSO4 and NaOH Lets look at that last one a bit closerβ¦. precipitate You can calculate if this will happen Q<K no precipitate Q=K saturated solution Q>K precipitate Spectator ions Basically same concept as Le Chatlierβs Priciple and shifting right or left! From earlier chapters Precipitation Examples If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0 mL of 0.10M Na2CO3 will BaCO3 precipitate? Ksp=8.1x10-9 Toughy: A volume of 75 mL of 0.060 M NaF is mixed with 25 mL of 0.15M Sr(NO3)2. Calculate the concentrations in the final solution of NO3- , Na+, and Sr2+, F-. (Ksp for SrF2=2.0x10-10) Review Two solutions with soluble ions can form a precipitate if mixed. A precipitate is formed if any combination of the ions are a slightly soluble compound AND if the Q>K If a precipitate is formed you can calculate how much is formed using stoichiometry. (limiting reagent problem) You can figure out how much of each ion is present. Depending on the ion: Treat as spectator ion Excess reagent Or find using Ksp Selective Precipitation and precipitating agents Learning Outcomes Use a selective precipitation scheme, and results from an experiment based on it to determine which ions are present in a sample. (note: do not memorize these schemes, they will be given to you). Calculate the amount of a precipitating agent needed to precipitate ions at a given concentration. Use the above knowledge to tell how pure compounds precipitated by selective precipitation will be. Selective Precipitation Unknown 1: Step 1: Add peroxide and NaOH. No precipitate present. Step 2: Add HNO3 and NH3: precipitate present. Al(OH)3 Selective Precipitation Unknown 2: Step 1: Add peroxide and NaOH. Precipitate present. Step 2: Add H2SO4 precipitate present. Step 3: HNO3 and NaBiO3 Step 4: Add KNO3, precipitate K3(Co(NO2)6) Co2+ Selective Precipitation: Difference in solubility Two ions with different Ksp can be separated by slowly adding a precipitating agent: Example: BaSO4 has a Ksp of 1.1x10-10 PbSO4 has a Ksp of 1.6x10-6 If a solution has dissolved barium and lead ions, which will precipitate first if a small amount of sulfate is added? Lower Ksp= lower solubility =BaSO4 Example: Selective Precipitation: Difference in solubility BaSO4 has a Ksp of 1.1x10-10 PbSO4 has a Ksp of 1.6x10-6 At what concentration of sulfate would barium start to precipitate if the solution contained 0.010M barium ions? At what concentration of sulfate would lead start to precipitate if the solution contained 0.010 M lead ions? πΎπ π = π΅π2+ [ππ42β ] [ππ42β ] πΎπ π = π΅π2+ [ππ42β ] = [ππ42β ] = 1.1π₯10β10 0.010π 1.1π₯10β8 M πΎπ π = ππ 2+ [ππ42β ] [ππ42β ] πΎπ π = ππ 2+ β6 1.6π₯10 [ππ42β ] = 0.010π [ππ42β ] = 1.6π₯10β6 M Example: A chemist attempts to separate barium ions from lead ions by using the sulfate ion as a precipitating agent. What is the concentration of barium ions when the lead sulfate begins to precipitate? 1) You need to find what sulfate concentration will precipitate each ion. We did this on the last slide ο! Lead Barium [ππ42β ] = 1.1π₯10β8 M [ππ42β ] = 1.6π₯10β6 M 2) What do we know about barium when lead starts to precipitate? We donβt! BUT we know the concentration of sulfate at that moment in time. SOβ¦..Use the sulfate when LEAD starts to precipitate to calculate the [Ba2+] concentration from the BaSO4 Ksp. πΎπ π = π΅π2+ [ππ42β ] π΅π2+ πΎπ π = [ππ42β ] π΅π2+ 1.1π₯10β10 β6 π = = 6.9π₯10 1.6π₯10β6 M Review Selective precipitation can be used when two ions have different solubility rules where one precipitates and one doesnβt. There are schemes worked out to separate out lots of these ions based on their solubility rules. You can also separate out ions that both precipitate with the same counter ion, BUT have different Ksp. Remember though there will be some contamination. The further away the Ksp are from each other, the less cross contamination. Ksp and tooth decay. Learning outcomes Combine materials from all chapters to: Describe the reasons behind common advice given by dentists about diet for preventing tooth decay. Define how common tooth decay preventions work. And yes, these are all based on solubility and acid/base chemistry!!!! Tooth Decay How does teeth decay occur? How does your body naturally prevent tooth decay? Why does sugar increase tooth decay? How do we artificially help prevent tooth decay and how do these processes work? How does teeth decay occur? Tooth Decay How does teeth decay occur? How does your body naturally prevent tooth decay? Why does sugar increase tooth decay? How do we artificially help prevent tooth decay and how do these processes work? How does your body naturally prevent tooth decay? Saliva! PO43- and HPO42- Tooth Decay Why does sugar increase tooth decay? Bacteria create acidic metabolites! Tooth Decay How do we artificially help prevent tooth decay and how do these processes work? Tooth Brushing: Fluoride Treatments Sugar Free Gum: Tooth Decay How do we artificially help prevent tooth decay and how do these processes work? Tooth Brushing: removes food Fluoride Treatments Sugar Free Gum: Increases Saliva Review Tooth decay is generally caused by acidic metabolites from bacteria eroding enamel. Important general concept: For things that react with acidic compounds, acid increases solubility. We combat not only by physically removing the food and bacteria but altering the chemistry Gum (sugar free) causes increased saliva which has buffering/basic compounds. Important general concept: buffers decrease the effect of acids Fluoride treatments change the enamel of our teeth to a compound with a lower Ksp. Important general concept: lower Ksp means compounds are less soluble. Complex Ion formation. Note: We are running behind so Iβm not going to cover this in class, you will not have homework on it and you will not be tested on it. However I will include my notes on it in case you are curious about what I have to say on it. This is solely FYI. I am happy to answer questions on it in office hours/facebook/through the survey if you have any. Complex Ion Formation. Used to increase the solubility of a salt. Using a second reaction. E.G. How does this work? βhidesβ ions away, allowing for more to be dissolved. What is the new solubility of AgCl? Lets add reactions and find the new K Steps for solving complex ion formation problems: Step 1: Find K Use Ksp and Kf Step 2: Use K to solve solubility problem as previous Ksp problems ICE chart (if you want) Fill into K equation. Complex Ion Formation: Example Calculate the molar solubility of AgCl in pure water and 3.0 M NH3 First Pure water: Complex Ion Formation: Example Calculate the molar solubility of AgCl in pure water and 3.0 M NH3 3.0 M NH3