Chemistry 104 Chapter 17 Summary and Review Answers

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1 Chemistry 104 Chapter 17 Summary ad Review Aswers 1. You eed to make a buffer with a ph of You have M sodium hydroxide, M acetic acid ad M sodium bicarboate. What do you do? (Make 50 ml total of the buffer.) Does the ph icrease or decrease with a small additio of hydrochloric acid? Aswer: I order to make a buffer with a ph of 4.80, we will eed to use the weak species (weak acid, as the ph is less tha 7), that will have a pk a approximately equal to the ph. Therefore, we have the choices of: Acetic acid (CH 3 COOH) K a = ; pk a = 4.74 Carboic acid (H 2 CO 3 ) K a1 = ; pk a = 6.38 K a2 = ; pk a = It might appear that you caot make a buffer regardless of havig the correct pk a. However, let s evaluate the eutralizatio of acetic acid with sodium hydroxide: CH COOH aq OH aq CH COO aq H O l So, if you add some sodium hydroxide to acetic acid, you will eutralize oly part of the weak acid (acetic acid) also formig the acetate io givig us a buffer. Now we eed to calculate the amout of each solutio (NaOH ad CH 3 COOH) that we eed: CH3COOH aq OH aq CH3COO aq H2O l I acid strog base 0 C strog base strog base strog base E 0 acid strog base strog base From the Hederso Hasselbalch equatio, we ca calculate the eeded ratio of cocetratio of the weak base to the weak acid: base base strog base phpkalog pka log log 1.15 acid strog base acid acid strog base acid strog base We also kow we are usig M acetic acid ad M sodium hydroxide (strog base), so we ca use this to determie volumes (of acid ad strog base): strog base acid Vstrog base Vacid M M Lastly, we kow the total volume is 50.0 ml: V acid + V strog base = L Now, solvig the four equatios for the variables of umber of moles: acid strog base 1.15 acid acid 1.15 acid strog base mol mol acid acid strog base

2 The volumes of acid/strog base would the be calculated: strog base mol acid mol Vstrog base 17.4 ml Vacid 32.5 ml M M M M Fially, we ca check back to see if our calculatios were correct: CH3COOHaq OH aqch3coo aqh2o l I C E base base phpka log pka log 4.74 log 4.80 acid Check! acid If HCl was added, the ph would decrease (ad the eutralizatio reactio would be CH 3 COO (aq) + H + (aq) CH 3 COOH) If you eeded a buffer with a ph of 6.50, what would you do? Do you see the process to use carboic acid ad sodium hydroxide? 2. You are titratig ml of M ethyl amie with M itric acid. What is the ph whe: a. No acid has bee added. b. Whe 13.0 ml of the acid has bee added. c. At the equivalece poit (what volume of the acid has bee added?) d. Whe 30.0 ml of the acid has bee added. Plot this ad idetify the equivalece poit ad the buffer regio. Aswer: These are the four typical locatios of calculatios for a titratio. All of these ivolve eutralizatio reactios except the first which is a equilibrium (o titrat or strog acid added yet). Part a: This is the iitial poit. For this equilibrium, we will eed the K b of CH 3 CH 2 NH 2 ad the cocetratio of the weak base. K b (CH 3 CH 2 NH 2 ) = (Table 16.5, p. 570 of your text) CH3CH2NH2 aq H2O l CH3CH2NH 3 aq OH aq I M 0 0 C x x x E x x x OH CH 2 3CH2NH 3 x Kb CH CH2NH x 2 4 Kb CH CH NH caot approximate

3 Kb Kb Kb x M= OH 2 poh = log( ) = ph = 14 poh = Part b: This is ow i the buffer regio. We will see that oce we complete the eutralizatio because oly part of the weak base is eutralized with the strog acid, leavig appreciable quatities of the weak base ad its cojugate acid (to solve this, we will use moles which we will calculate first): CH3CH2NH2 H mol ml mol L mol ml mol L CH CH NH aq H aq CH CH NH aq I C E V CH3CH2NH2 H Now we ca use the Hederso Hasselbalch to solve for the ph (rememberig we have the K b value ad pk a + pk b = 14): base base phpkalog 14 pkblog log acid Part c: This is ow at the equivalece poit. This meas that the umber of moles of acid equals the umber of moles of base. We will see that oce we complete the eutralizatio all of the weak base is eutralized with the strog acid, leavig appreciable quatities of the cojugate acid (the requirig aother equilibrium calculatio to determie the ph): mol H H mol 28.8 ml H M acid CH3CH2NH2 aq H aq CH3CH2NH 3 aq I C E Cocetratio of the weak cojugate acid is: CH3CH2NH mol 3 CH3CH2NH M V L L T Now we agai cosider equilibrium (but ow of the weak acid): CH CH NH aq CH CH NH aq H aq I C x x x E x x x H CH 2 3CH2NH 2 x Ka CH CH2NH x 3 11 K CH CH NH a

4 2 2 x x Ka x Ka x CH3CH2NH2 H ph log H 5.96 Part d: This is ow i the excess regio. This meas that the umber of moles of acid has exceeded the umber of moles of base ad the ph is drive by the excess strog acid. We will see that oce we complete the eutralizatio all of the weak base is eutralized with excess strog acid remaiig: mol mol ml mol L H CH3CH2NH2 aq H aq CH3CH2NH 3 aq I C E mol M p log H log H H VT L L Now, combiig these four poits ito the titratio curve, we will plot ph versus volume of the strog acid (HNO 3 ) added. All four phs that we calculated are oted below correspodig to the four calculatios above. Iitial poit ph = ph = 5.96 This is the equivalece poit. At this poit, 28.8 ml of the acid has bee added ph ph = This is the buffer regio (where ph pk a ) ph = This is the excess regio. Volume of HNO 3 added (i ml)

5 3. Desig a scheme to defiitively idetify the presece of oe or more of the catios, Pb 2+,,, Fe 2+, Ag +, ad. To do this you must assume that the sample could cotai all 7 catios. Your reagets are: Na 2 CO 3 (aq), HCl(aq), NaOH(aq), KSCN(aq), Na 2 CrO 4 (aq), Na 2 SO 4 (aq), H 2 S(aq) ad NaF(aq) Your scheme must be clear eough that each precipitatio is quatitative (all of a particular catio tested for is removed by the test so as ot to iterfere with further tests). This ca be doe easily by usig the commo io effect (add excess reaget). Your scheme must also be sesitive to the order of the additio of reagets (i.e. remove the most isoluble catio first why?). We are goig to use a series of K sp values give here as referece: Solubility Product Costat, K sp, Table Carboates K sp Chromates K sp Sulfides K sp BaCO BaCrO FeS CaCO CaCrO PbS MgCO PbCrO Ag 2 S Ag 2 CrO Chlorides K sp Sulfates K sp Thiocyaates K sp AgCl BaSO AgSCN PbCl CaSO PbSO Hydroxides K sp Fluorides K sp Al(OH) MgF Fe(OH) CaF Pb(OH) PbF AgOH Mg(OH) Aswer: There may have bee two ways that this problem was read. Versio 1 Step 1: Simply devise a scheme which ca idetify the presece of a io (without questio o overlap with aother precipitatio). Add NaSCN Ba2+ Fe 2+ Ag + Pb 2+ AgSCN

6 Add HCl Step 2: Fe 2+ Pb 2+ PbCl 2 Step 3: Add H 2 S Fe Mg FeS Now thigs are a little tricky. Addig oe io to precipitate oe catio also precipitates aother. Therefore, the remaiig catios (remember, we have ow removed as solids AgSCN, PbCl 2, ad FeS) are separated ito three cotaiers for further aalysis. Step 4: Add Na 2 CO 3 The add NaOH CaCO 3 BaCO 3 MgCO 3 Al(OH) 3 Step 5: Add NaOH The add NaF Al(OH) 3 Mg(OH) 2 CaF 2 The add NaSO 4 BaSO 4

7 Step 6: Add Na 2 CrO 4 The add NaF BaCrO 4 CaCrO 4 MgF 2 Versio 2 Now devise a scheme which ca idetify the presece of a io defiitively ad selectively precipitate all of the io (with o overlap). This is a little more tricky but doable. Step 1 through 3 are the same. Step 4: Add a small Na 2 CrO 4 Add excess Na 2 CrO 4 BaCrO 4 CaCrO 4 How much is small? To aswer this, determie the miimum cocetratio of chromate to precipitate barium chromate versus calcium chromate. (We will assume the cocetratio of the metal ios is M ad the additio of cocetrated sodium chromate will ot chage the total volume so the cocetratio of the metal ios i solutio is costat at M uless removed by precipitatio.) BaCrO 4 : Ksp Ba CrO M CrO 4 CrO M ad greater tha this cocetratio precipitates BaCrO CaCrO 4 : Ksp Ca CrO M CrO 4 CrO M ad greater tha this cocetratio precipitates CaCrO

8 So small amout is aythig less tha M. But, you may ask, wo t there still be barium ios i solutio whe calcium chromate begis to precipitate? (Good questio) If you calculate the cocetratio of barium ios i solutio whe the cocetratio of chromate has almost reached the cocetratio to precipitate calcium chromate: Ksp Ba CrO 4 Ba M Ba M % of the barium remais i solutio! Step 5: Add NaF Add NaOH MgF 2 Al(OH) 3