Semester 2 AP Chemistry Unit 12

Transcription

1 Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved salt greatly inhibits the dissciatin f the acid The psitin f the dissciatin equilibrium has been shifted left by the presence f the ins frm the salt What is a Buffer? A buffer slutin may cntain... A weak acid and its salt Ex) HF and NaF A weak base and its salt Ex) NH 3 and NH 4 Cl The slutin has the ability t resist change in ph upn the additin f small amunts f acid r base The mst practical example f a buffer slutin is ut bld, it absrbs acids and bases withut changing its ph ph f a buffer slutin A buffered slutin cntains 0.50 M acetic acid and 0.50 M sdium acetate. Calculate the ph f the slutin. Acetic Acid H+ + Acetate I 0.50 M C -x +x +x E 0.50 x x 0.50+x Slve using the Ka Value per usual The acetic acid dissciatin equilibrium will cntrl the ph f the slutin Adding a base t Buffer Slutin 1) Assume the reactin ges t cmpletin and use stichimetric ratis YOU MUST CONVERT TO MOLES!!!! 2) Use the riginal equatin t carry ut the equilibrium calculatins CONVERT BACK TO MOLARITY!!! Vlumes are additive! Adding an acid t Buffer slutin 1) Assume the reactin ges t cmpletin and use stichimetric ratis MUST CONVERT TO MOLES!!!! 2) Use the riginal equatin t carry ut the equilibrium calculatins CONVERT BACK TO MOLARITY!!! Vlumes are additive! Hendersn-Hasselbalch The equatin is useful fr calculatin the ph f the slutin when the rati f [A-] t [HA] is knwn

2 cnjugate base Frmula: ph = pka + lg acid Buffer Slutins with Weak Bases B + H 2 O BH + + OH 1) The Weak base reacts with any H+ added 2) And the cnjugate acid reacts with any added OH- Cnvert mles back t mlarity and insert int the riginal equatin Alternatively We can btain the Ka value fr Ammnium frm the given Kb value fr NH3 Frmula: poh = pkb + lg cnjuagte acid Types f Buffer Slutins Type 1: Cmbine a weak acid with its cnjugate base Ex) HF and KF (a basic salt) Type 2: Cmbine a weak base with its cnjugate acid Ex) NH3 and NH4Cl (an acidic salt) Remember that NH3 can als be written as NH4OH Type 3: Add a limited amunt f a strng acid with an excess f a weak base. The result will be the same slutin as in type 2 Ex) add 0.70 mles f HCl t 1.0 Liter f a 2.0 M slutin f NH3. This will result in an excess f 1.30 mles f the base and generating 0.70 mles f the NH4 in Type 4: Add a limiting amunt f a strng base with an excess f a weak acid. The result will be the same as in type 1 Ex) add 0.50 mles f NaOH t 1.0 Liter f a 2.0 M slutin f HF. This will result in an excess f 1.5 mles f the acid and generating 0.50 mles f the F- in. Type 5: Prepare a slutin f an amphteric substance Ex) 1.0 M slutin f NaHCO3 Preparing a buffer: Resisting change lab If the mlar cncentratin f the acid and its cnjugate base are abut equal then the rati wuld be 1.0 and the ph wuld equal the pka Steps: 1) Chse a weak acid whse pka is clse t the ph we want, abut + r 1 2) Substitute the ph and the pka values int the H-H equatin 3) Slve fr the [Cnjugate Base] t [acid] rati 4) This rati can then be cnverted t mlar quantities t prepare a buffer Slubility Equilibria This cncept invlves salts f lw slubility A saturated slutin f calcium phsphate is in cntact with slid calcium phsphate base

3 Only very small amunts f calcium phsphate will disslve in water What des, will dissciate cmpletely int calcium ins and phsphate ins Therefre, an equilibrium will be established between undisslved calcium phsphate and its ins in slutin Equilibrium cnstant Ksp is called the slubility prduct cnstant r simply the slubility prduct The slubility prduct f a cmpund in the prduct f the mlar cncentratin f the cnstituent ins, each raised t the pwer f its stichimetric cefficient in the balanced equilibrium equatin Frm unit 10 we knw that fr hetergeneus equilibria, the cncentratin f a slid r liquid is cnstant High temperature = mre disslved/mre sluble Except fr gases Mlar slubility The number f mles f slute in 1 Liter f a saturated slutin (ml/l) Cncentratins f the ins in the slubility prduct expressin are als mlar cncentratins Prblem Type 1: If the slubility f Fe(OH)2 in water is 7.7*10^-8 M at a certain temperature what is its Ksp value at that temperature? 1) Write the equilibrium reactin and then the slubility prduct expressin 2) Cncentratins can be fund frm the stichimetric ratins fund in the equilibrium equatin 3) Substitute the equilibrium cncentratins int the slubility prduct expressin t calculate Ksp A substance s slubility can be expressing in tw ways Mlar slubility (ml/l) Slubility (grams f slute/1l f a saturated slutin) Prblem Type 2: The slubility f calcium sulfate is fund experimentally t be 0.67g/L. Calculate the value f Ksp fr calcium sulfate. 1) Write the equilibrium reactin and then the slubility prduct expressin 2) Using the slubility, calculate the mlar cncentratins f each in fund in sultin 3) Substitute the equilibrium cncentratins int the slubility prduct expressin t calculate Ksp Smetimes we are given the calue f Ksp fr a cmpund and asked t calculate the cmpund s mlar slubility The expected cncentratins at equilibrium can be calculated frm the initial cncentratins and the equilibrium cnstant Equilibrium cncentratins = initial cncentratin +/- the change due t reactin Prblem Type 3: What is the mlar slubility f silver phsphate in water?

4 1) Write the equilibrium reactin and then the slubility prduct expressin 2) A small amunt f silver phsphate will dissciate in slutin. Represent this amunt as s. Make an ICE chart!!! since ne unit f silver phsphate yields three silver ins and ne phsphate in, at equilibrium the cncentratin f silver is +3s and the cncentratin f phsphate is +s 3) Substitute the Ksp and the cncentratins f silver and phsphate in terms f s int the slubility prduct expressin t slve fr s, the mlar slubility Predicating a precipitatin reactin T predict whether a precipitate will frm, we must calculate the in prduct Q The in prduct represents the mlar cncentratins f the ins raised t the pwer f their stichimetric cefficients The subscript 0 indicates that these are initial cncentratins and d nt necessarily crrespnd t equilibrium cncentratins Q = Ksp The reactin is already at equilibrium, the slutin is saturated Q < Ksp The slutin is nt saturated Mre slid culd disslve t prduce mre ins in slutin N precipitate will frm Q > Ksp The slutin is supersaturated There are t many ins in slutin Sme ins will cmbine t frm a precipitate until the prduct f the in cncentratins is equal t the Ksp Prblem Type 4: Predict whether a precipitate f PbI will frm when 200 ml f M Pb(NO3)2 and 300 ml f M NaI are mixed tgether 1. Calculate the cncentratins f the cmpunds after mixing the tw slutins tgether. This is a dilute prblem. MV=MV 2. Substitute the cnentratins int the in prudcut expressin t calculate Q, then cmpare Q and Ksp t determine if a precipitate will frm Effect f a Cmmn In n Slubility Cnsider the slightly sluble salt Cpper (I) Idide. Suppse we disslve CuI in a ptassium idide slutin. Wuld the presence f KI affect the slubility f CuI? KI will inize cmpletely The I- ins frm KI will affect the CuI equilibrium

5 The additin will shift the reactin t the left t remve sme f the additinal I- ins, decreasing the slubility f CuI When calculating the slubility f a slat (in the presence f a cmmn in) the nly difference is the initial cncentratin f the cmmn in will nt be zer Prblem Type 5: What is the mlar slubility f PbCl2 in a 0.50 M NaCl slutin? 1. Write ut the equilibrium reactin. Write the slubility prduct expressin 2. ICE CHART!!! 3. Substitute the value f Ksp and the cncentratin f the ins int the slubility cnstant expressin ph and Slubility The slubilities f many substances als depend n the ph f the slutin Cnsider the fllwing equilibrium Mg(OH) 2(s) Mg OH Adding OH- ins (increasing ph) shifts t the left Adding H+ ins (decrease ph) shifts right and increases slubility and therefre insluble bases tend t disslve in acidic slutins and insluble acids tend t disslve in basic slutins