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Final Exam Wednesday, December 17 7:00 – 10:00 pm 100 Noyes Lab AQB,AQ1Davis AQ2,AQ3Roberts AQ5,AQ6Tsai 150 Animal Science AQ7,AQ8Vuong AQJ Livingston 217 Noyes Lab AQA,AQCChang 192 Lincoln AQ4,AQ9Patel
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Free Tutoring! Chem 101, Chem 102, Chem 103, Chem 104 Econ 102, Econ 103 Math 221, Math 231, Math 234 PHYS 211 Stop by GREGORY HALL ROOM 327 AND 329 FROM 2-4PM on December 11 for free tutoring from Phi Eta Sigma Questions? PEStutoring2@gmail.comPEStutoring2@gmail.com
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Final Exam Friday, December 19 1:30 – 4:30 pm 100 Noyes Lab EQ2,EQ3,EQC,EQ8Steele EQ4Ip 141 Wohler EQ1,EQ6,EQ7,EQBSmall EQ5Ip
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Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log H + = 1.00.
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. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL10.0 mL 2.5 x 10 -3 mol1.0x 10 -3 mol-= 1.5 x 10 -3 mol V = 25 + 10 mL [H + ] =1.5 x 10 -3 mol 35 x 10 -3 L [H + ] = 4.28 x 10 -2 M pH =1.37.
![. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL10.0 mL 2.5 x mol1.0x mol-= 1.5 x mol V = mL [H + ] =1.5 x mol 35 x L [H + ] = 4.28 x M pH =1.37.](http://images.slideplayer.com./26/8401688/slides/slide_5.jpg ". Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL10.0 mL 2.5 x mol1.0x mol-= 1.5 x mol V = mL [H + ] =1.5 x mol 35 x L [H + ] = 4.28 x M pH =1.37.")
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. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol. 20.0 mL 2.0 x 10 -3 mol-= 0.5 x 10 -3 mol V = 25 + 20 mL [H + ] =0.5 x 10 -3 mol 45 x 10 -3 L [H + ] = 1.11 x 10 -2 M pH =1.95.
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. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol. 25.0 mL 2.5 x 10 -3 mol-= 0.0 mol [H + ] = pH =7.00. 1.00 x 10 -7 M.
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. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol. 35.0 mL V = 25 + 35 mL [OH - ] =1.0 x 10 -3 mol 60 x 10 -3 L [OH-] = 1.67 x 10 -2 M pOH =1.78. 3.5 x 10 -3 mol OH -. pH = 12.22.
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Titration Curves Weak Acid+ Strong Base 0.1 M CH 3 COOH0.1 M NaOH 25.0 mL Initialweak acid half-way pointpH = pK a K a = 1.8 x 10 -5 equivalence point = [H + ] [CH 3 COO - ] [CH 3 COOH] = 4.74 CH 3 COO - + H 2 O CH 3 COOH+ OH - K b = 5.6 x 10 -10 = [OH - ] [CH 3 COOH] [CH 3 COO - ] strong base pH = pK a + log [CH 3 COO - ] [CH 3 COOH]
![Titration Curves Weak Acid+ Strong Base 0.1 M CH 3 COOH0.1 M NaOH 25.0 mL Initialweak acid half-way pointpH = pK a K a = 1.8 x equivalence point = [H + ] [CH 3 COO - ] [CH 3 COOH] = 4.74 CH 3 COO - + H 2 O CH 3 COOH+ OH - K b = 5.6 x = [OH - ] [CH 3 COOH] [CH 3 COO - ] strong base pH = pK a + log [CH 3 COO - ] [CH 3 COOH]](http://images.slideplayer.com./26/8401688/slides/slide_9.jpg "Titration Curves Weak Acid+ Strong Base 0.1 M CH 3 COOH0.1 M NaOH 25.0 mL Initialweak acid half-way pointpH = pK a K a = 1.8 x equivalence point = [H + ] [CH 3 COO - ] [CH 3 COOH] = 4.74 CH 3 COO - + H 2 O CH 3 COOH+ OH - K b = 5.6 x = [OH - ] [CH 3 COOH] [CH 3 COO - ] strong base pH = pK a + log [CH 3 COO - ] [CH 3 COOH]")
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Titration Curves Weak Base+ Strong Acid 0.1 M NH 3 0.1 M HCl 25.0 mL 1. Initial pH NH 3 NH 4 + + OH - [OH - ] = 1.34 x 10 -3 M pOH =2.87 Kb=Kb=1.8 x 10 -5 =[OH - ][NH 4 + ] [NH 3 ] + H 2 O pH = 11.12
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Titration Curves Weak Base+ Strong Acid 0.1 M NH 3 0.1 M HCl 25.0 mL 2.5 x 10 -3 mol NH 3 NH 4 + + OH - Kb=Kb=1.8 x 10 -5 =[NH 4 + ][OH - ] [NH 3 ] 10.0 mL 1.0 x 10 -3 mol-= 1.5 x 10 -3 mol V = 25 + 10 mL [NH 3 ] [NH 4 +] [OH - ] 0.0430.0290.0 0.043 -x0.029 + xx 1.8 x 10 -5 =[x][0.029 + x] [0.043 - x] x = 2.67 x 10 -5 pOH = 4.57 + H 2 O pH = 9.43.
![Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol NH 3 NH OH - Kb=Kb=1.8 x =[NH 4 + ][OH - ] [NH 3 ] 10.0 mL 1.0 x mol-= 1.5 x mol V = mL [NH 3 ] [NH 4 +] [OH - ] x xx 1.8 x =[x][ x] [ x] x = 2.67 x pOH = H 2 O pH = 9.43.](http://images.slideplayer.com./26/8401688/slides/slide_11.jpg "Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol NH 3 NH OH - Kb=Kb=1.8 x =[NH 4 + ][OH - ] [NH 3 ] 10.0 mL 1.0 x mol-= 1.5 x mol V = mL [NH 3 ] [NH 4 +] [OH - ] x xx 1.8 x =[x][ x] [ x] x = 2.67 x pOH = H 2 O pH = 9.43.")
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Titration Curves Weak Base+ Strong Acid 0.1 M NH 3 0.1 M HCl 25.0 mL 2.5 x 10 -3 mol NH 3 NH 4 + + OH - Kb=Kb=1.8 x 10 -5 =[NH 4 + ][OH - ] [NH 3 ] 20.0 mL 2.0 x 10 -3 mol-= 5.0 x 10 -4 mol V = 25 + 20 mL [NH 3 ] [NH 4 +] [OH - ] 0.0110.0440.0 0.011 -x0.044 + xx 1.8 x 10 -5 =[x][0.044 + x] [0.011 - x] x = 4.5 x 10 -6 pOH = 5.35 + H 2 O pH = 8.65..
![Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol NH 3 NH OH - Kb=Kb=1.8 x =[NH 4 + ][OH - ] [NH 3 ] 20.0 mL 2.0 x mol-= 5.0 x mol V = mL [NH 3 ] [NH 4 +] [OH - ] x xx 1.8 x =[x][ x] [ x] x = 4.5 x pOH = H 2 O pH =](http://images.slideplayer.com./26/8401688/slides/slide_12.jpg "Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol NH 3 NH OH - Kb=Kb=1.8 x =[NH 4 + ][OH - ] [NH 3 ] 20.0 mL 2.0 x mol-= 5.0 x mol V = mL [NH 3 ] [NH 4 +] [OH - ] x xx 1.8 x =[x][ x] [ x] x = 4.5 x pOH = H 2 O pH =")
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Titration Curves Weak Base+ Strong Acid 0.1 M NH 3 0.1 M HCl 25.0 mL 2.5 x 10 -3 mol NH 3 NH 4 + +H + Ka=Ka=5.6 x 10 -10 =[NH 3 ][H + ] [NH 4 + ] 25.0 mL 2.5 x 10 -3 mol-= 0.00 V = 25 + 25 mL [NH 4 ] [NH 3 ] [H + ] 0.050.000.0 0.05 -x xx 5.6 x 10 -10 =[x 2 ] [0.05 - x] x = 5.9 x 10 -6 pH = 5.27...
![Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol NH 3 NH 4 + +H + Ka=Ka=5.6 x =[NH 3 ][H + ] [NH 4 + ] 25.0 mL 2.5 x mol-= 0.00 V = mL [NH 4 ] [NH 3 ] [H + ] x xx 5.6 x =[x 2 ] [ x] x = 5.9 x pH =](http://images.slideplayer.com./26/8401688/slides/slide_13.jpg "Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol NH 3 NH 4 + +H + Ka=Ka=5.6 x =[NH 3 ][H + ] [NH 4 + ] 25.0 mL 2.5 x mol-= 0.00 V = mL [NH 4 ] [NH 3 ] [H + ] x xx 5.6 x =[x 2 ] [ x] x = 5.9 x pH =")
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Titration Curves Weak Base+ Strong Acid 0.1 M NH 3 0.1 M HCl 25.0 mL 2.5 x 10 -3 mol 20.0 mL pOH = pK b + log [NH 4 + ] [NH 3 ] 5.0 x 10 -4 mol NH 3 2.0 x 10 -3 mol NH 4 + V = 45 x 10 -3 L pOH = K a = 1.8 x 10 -5 4.74+ log (0.44) (0.11) = 5.34 pH = 8.65
![Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol 20.0 mL pOH = pK b + log [NH 4 + ] [NH 3 ] 5.0 x mol NH x mol NH 4 + V = 45 x L pOH = K a = 1.8 x log (0.44) (0.11) = 5.34 pH = 8.65](http://images.slideplayer.com./26/8401688/slides/slide_14.jpg "Titration Curves Weak Base+ Strong Acid 0.1 M NH M HCl 25.0 mL 2.5 x mol 20.0 mL pOH = pK b + log [NH 4 + ] [NH 3 ] 5.0 x mol NH x mol NH 4 + V = 45 x L pOH = K a = 1.8 x log (0.44) (0.11) = 5.34 pH = 8.65")
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Polyprotic Acid H 2 SO 3 HSO 3 - + H + K a1 = 1.4 x 10 -2 HSO 3 - SO 3 2- + H + K a2 = 6.5 x 10 -8 2 equivalents of base 0.10 M H 2 SO 3 0.10 M NaOH 40 mL80 mL Initial pH 1.4 x 10 -2 =[HSO 3 - ][H + ] [H 2 SO 3 ] = x 2 0.1 - x x = 0.03pH = 1.51
![Polyprotic Acid H 2 SO 3 HSO H + K a1 = 1.4 x HSO 3 - SO H + K a2 = 6.5 x equivalents of base 0.10 M H 2 SO M NaOH 40 mL80 mL Initial pH 1.4 x =[HSO 3 - ][H + ] [H 2 SO 3 ] = x x x = 0.03pH = 1.51](http://images.slideplayer.com./26/8401688/slides/slide_15.jpg "Polyprotic Acid H 2 SO 3 HSO H + K a1 = 1.4 x HSO 3 - SO H + K a2 = 6.5 x equivalents of base 0.10 M H 2 SO M NaOH 40 mL80 mL Initial pH 1.4 x =[HSO 3 - ][H + ] [H 2 SO 3 ] = x x x = 0.03pH = 1.51")
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Polyprotic Acid H 2 SO 3 HSO 3 - + H + K a1 = 1.4 x 10 -2 HSO 3 - SO 3 2- + H + K a2 = 6.5 x 10 -8 2 equivalents of base 0.10 M H 2 SO 3 0.1 M NaOH half-way point pH = pK a - log 1.4 x 10 -2 =1.85. - log 6.5 x 10 -8 =7.19. 1 st equivalence point 1.84 + 7.19 2 = 4.52. 2 nd equivalence point conjugate base, SO 3 -
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Polyprotic Acid H 2 SO 3 HSO 3 - + H + K a1 = 1.4 x 10 -2 HSO 3 - SO 3 2- + H + K a2 = 6.5 x 10 -8 2 equivalents of base 0.10 M H 2 SO 3 half-way point pH = pK a - log 1.4 x 10 -2 =1.85. - log 6.5 x 10 -8 =7.19. 1 st equivalence point 1.84 + 7.19 2 = 4.52. 2 nd equivalence point conjugate base, SO 3 -. buffering regions K b2 = 1 x 10 -14 / 6.5 x 10 -8 = 1.53 x 10 -7 [SO 3 2- ] = 0.1 mol.120 L = 0.83 1.53 x 10 -7 = x 2 / 0.83 x = 3.57 x 10 -4 = [OH - ] pH = 10.44
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