Common Ion Effect: Makeup Midquarter Exams Wed., Mar 9 5:30-7:30 pm 131 Hitchcock Hall. You MUST Sign up in 100 CE. Please do so as soon as possible.

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1 Makeup Miquarter Eams We., Mar 9 5:0-7:0 pm 11 Hitchcock Hall Chapter 17 Aitional Aspects of Aqueous Equilibria You MUST Sign up in 100 CE Please o so as soon as possible. Final Eams for Chem Mathews 17.1 The Common-on Effect 17.2 Buffere Solutions Composition an Action of Buffere Solutions Buffer Capacity an ph Aition of Strong Acis or Bases to Buffers 17. Aci-Base Titrations Strong Aci-Strong Base Titrations Weak Aci-Strong Base Titrations Titrations of Polyprotic Acis 2:0 Class: Thursay, Mar. 17 at 1:0 in H :0 Class: Tuesay, Mar. 15 at 5:0pm in MP 1000 Common on Effect: This is the same as other uilibrium calculations, ecept there is an initial concentration of Consier a solution which is prepare by miing 0.10 mole of acetic aci an mole of HCl in enough water to make 0 L of solution. What are the concentrations of each species? HOAc 0.10 H O 0.01 OAc - two species prior to establishing uilibrium Note that for a solution of 0.10 M acetic aci alone, [OAc - = [H O = M Consier a solution mae from issolving 0.0 moles of acetic aci an 0.0 moles of soium acetate in enough water to make L of solution. What is the concentration of Hyronium ions an the ph? Consier L of solution containing mol of formic aci (HCOOH) an mol of soium formate (NaHCOO). What is the ph? K a = HOAc = H O OAc - i 0.0 M ~ M δ - HA - - H O A - (0.0-) (0.0) K a = = [()(0.0) / [(0.0-) leas to = [H O = M compare to = ()() / (-) ()() / () so = = [H O an ph = M for the aci alone!!!

2 Note, however, that we re oing the same things again! The epression Ka = [H O [A - / [HA Now let s a 0.10 mol of strong aci (eg HCl) to the solution (without changing its volume). HA H O A - can be written [H O = K a [HA / [A - K a [HA i / [A - i or take the negative log of both sies: log [H O = -logk a - log [HA i / [A - i a 0.10 or ph = pk a - log [HA i / [A - i rn ~ or ph = pk a log[a - / [HA Henerson-Hasselbalch Equation(s) Vali ONLY when [H O an [OH - are very small in comparison to [A - i an [HA i An ph =pk a log(0.400)/(1.10) =.75(-0.44) =.1=pH or [H O = A similar calculation results when we a enough strong base to the original buffer solution to make the solution 0.10 M in OH -. HA H O A a rn 0.90 ~ n this case, [H O = an ph =.58 Chapter 17 Aitional Aspects of Aqueous Equilibria Final Eams for Chem Mathews 17.1 The Common-on Effect 17.2 Buffere Solutions Composition an Action of Buffere Solutions Buffer Capacity an ph Aition of Strong Acis or Bases to Buffers 17. Aci-Base Titrations Strong Aci-Strong Base Titrations Weak Aci-Strong Base Titrations Titrations of Polyprotic Acis 2:0 Class: Thursay, Mar. 17 at 1:0 in H :0 Class: Tuesay, Mar. 15 at 5:0pm in MP 1000

3 Titration curves: Strong Aci Strong Base titrate ml of M HCl with M NaOH Recognize that if c 0 an V o = concentration an volume of initial aci(bas) an c t an V t = concentration an volume of base (aci) ae Then the uivalence point correspons to the conition c o V o = c t V e, where V e = V t when n aci = n base 1. V = 0 ml of base ae n(h 0 ) = (0.1000M)( L) = mol H 0 an ph = 2. V = 0 ml of NaOH ae (i.e. 0 < V < V e ) n(oh - ) = ( M)(0.000 L) = mol OH - which reacts with (neutralizes) the same amount of H0 i.e. n (H) = ( ) mol H 0 = mol H 0 remaining (.000 mol Na an Cl - ) an [H 0 = ( )/(Va Vb) = ( ) / (0.100 L) = M an ph = V = ml NaOH ae (i.e. V b = V e ) n(oh- ae) = ( M)( L) = mol which neutralizes the same amount (all) of H 0. The significant reaction that has taken place is H OH - therefore [OH- = [H 0 = M an ph = 7.00 Note also that [Na = [Cl - = ( mol) / ( L) = 0 M

4 SA SB, continue: 4. V = ml (i.e. V b > V e ) also, this is about one rop ecess Titration of Strong Aci with Strong Base easiest to treat by realizing that ml gave the result of previous calculation, i.e. neutralize the aci. then n (OH- ecess) = [( )mL[ M = mol OH - etra an [OH - = ( mol) / ( ) = M OH - so that poh = 4.60 an ph = 9.40 similar calculations can procee for all ecess base Weak Aci Strong Base ml M HOAc titrate with M NaOH Note the efinition of the uivalence point remains the same. 1. V = 0.00 ml NaOH usual problem of weak aci, with K a = n(hoac) = ( M)( L) = mol HOAc so [H 0 = M an ph = < V < Ve aci will be partially neutralize. Since OH- is a stronger base than water, HOAc OH - = OAc - K = 1/K b = K a / K w = >>1 eample, 0 ml of NaOH n(oh-) = ( M)(0.000 L) = mol OHn(HOAc) = ( M)( L) = mol HOAc so after all the OH- has reacte with HOAc, n(oac-) = n(oh- ae) = mol n(hoac remaining) = ( ) mol HOAc = mol HOAc 2. (continue) [OAc- = ( mol OAC-) / (Va Vb) = ( mol OAC-) / (0.100 L) = M OAcan [HOAc = ( ) / (0.100 L) = M HOAc Now we re able to calculate the ph of this buffer soln ph = pka log{[oac- /[HOAc} = 4.75 log{( ) / ( )} ph = 4.8 Notice also that when V b = ½ V a, [HA=[A- an ph=pk a. V = 100mL NaOH, V = V e WA-SB titration (cont) we now have n(naoac) = n(oac- = (0.1000M)(0.1000L) = mol OAcwhich is the same as having mol of NaOAc in soln [OAc- = (010-2 ) / ( L) = M As before, we can calculate the ph base on hyrolysis OAc- H2O = HOAc OH- K b = K w /K a = an ph = 8.7

5 4. V >100mL NaOH, i.e, V = V e WA-SB titration (cont) Titration of Weak Aci with Strong Base, V e = 50 ml just aing more OH- to the solution of NaOAc, results primarily in the [OH- being ominate by the NaOH being ae i.e. this portion of the titration curve looks very similar to the SA-SB titration curve Titration of Weak Aci with Strong Base, V e = 50 ml ph=pk ph=pk a a Anything wrong with this picture??? Discuss weak polyprotic acis, eg H 2 CO

6 Weak polyprotic acis: typical H 2 CO an H PO 4 H 2 CO = H O HCO - K a1 = HCO - = H O CO 2- K a2 = Eample: Aqueous sol n sat in CO 2 ; [H 2 CO = 0.04 M (1) [H O = [HCO - = ; [H 2 CO = 0.04 M (2) [H O = [HCO - = same ; [CO 2- = Eample: at ph = 10.00, [H O = CO Ka = = = CO O [ CO CO = Ka2 = = O Note also we can rewrite the epressions: CO Ka 1 = CO O 2 2 [ CO Ka2 = CO O We can calculate the fraction of any species, eg H 2 CO 2CO fraction 2CO = 2 CO CO [ CO 2 2CO CO fraction[ H 2CO = 2 2CO CO [ CO CO CO CO fraction[ H CO = = similarly for HCO - = 0.68 an CO 2- = 0.2 An, this coul be repeate for all ph values

7 100 ml of M H PO 4 titrate with NaOH pk a1 = 2.27, pk a2 = 7.21, pk a = 12.25