Anti-Semitism was evident in theprofessions
Antisemitism did not appear in wake of the war but was rather a precursor that allowed Hitlers antisemitic sentiments to easily penetrate the masses.
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ADVERTISING EVENTS: CONTENT AND SEMIOTICANALYSES OF CANNES FILM FESTIVAL POSTERS
Rhetorical Analysis Essay
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learn-us-east-1-prod-fleet02-xythos.content.blackboardcdn.com learn-us-east-1-prod-fleet02-xythos.content.blackboardcdn.com
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Class Exercise 2
3.6*10^-6 M KOH
pOH = -logCb
pH = 14.00 - pOH
8.56
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Class Exercise 3
4*10^-3 M Sr(OH)2
pOH = -log(2*Cb)
pH = 14.00 - pOH
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Class Exercise 1
2.7*10^-4 M HBr
pH = -logCha
3.57
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Case 1: C B > 10 -6 M; [OH-] = N*C B ; pOH = - log([OH- ]; pH = 14.00 – pOH
Exam 2
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Ex. Calculate the pH of 0.0500 M Ba(OH)
1.0500 M Ba(OH)2
pOH = -log(2Cb) = -log(20.0500 M)
pH = 14.00 - pOH
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Ex. What is the pH of 0.1 M NaOH?
pOH = -log[OH^-] = -logCb
pH = 14.00 - pOH
pOH = -logCb = -log(0.1 M) = 0.1
pH = 14.00 - 1.0 = 13.0
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[OH- ] and pOH
Similar to strong acid however the result is pOH
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2. Strong base.
exam 2
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Case 1: C HA > 10-6 M; [H+ ] = C HA; pH = -log(C HA).
Exam 2 Topic
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Summary
For any concentration above 10^-6 M, 4Kw << GO BACK TO NOTES
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Eliminate [OH-] and [A-] yields a quadratic equation in [H+ ]
[H^+]^2 - Cha[H^+] - Kw = 0
[H^+] = (Cah + (Cha^2+ 4Kw)^1/2)/2
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Strong acid with formal concentration C HA
1.0*10^-8 M HCl
H2O <--> H^+ + OH^-
Kw = [H^+] [OH^-]
[H^+] = [OH^-] + [Cl^-]
Cha = [Cl^-]
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Consider this solution
pH = -log[H^+] = -log(1.0x10^-8) = 8
Adding acids can not give you a base Therefore, you can no longer ignore H+
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E.g., What is the pH of 0.10 M HCl solution?
0.10 M HCl
pH = -log[H^+] = -logCh+ = 1.00
pH = -log[H^+] = -logCha = -1.00
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(b) C HA is the formal concentration of the acid; (c) C B is the formalconcentration of the base
0.10 M CH3COOH
0.10 M --> Cha
0.10 M NaOH
0.10 M --> Cba
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learn-us-east-1-prod-fleet02-xythos.content.blackboardcdn.com learn-us-east-1-prod-fleet02-xythos.content.blackboardcdn.com
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Solving these equations
Sub-case 1: - the total amount (i.e. formal concentration) of one species is given
Sub-case 2: COME BACK TO THIS
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Class exercise
[Ag^+] = 2 ( [CO3^2-] + [HCO3^-] + [H2CO3] )
2 goes in front of the CO3 products
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Ex. CaF2 sol'n
2 [Ca^2+] = [F^-] + [HF]
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Ex. CaF2 sol'n
NOT 1:1 Ratio!
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Ex. PbCl2 sol’n
PbCl2 <--> Pb^2+ + 2Cl^-
_ [Pb^2+] = _ [Cl^-]
2x [Pb^2+] = _ [Cl^-] NOTICE: the 2x from Cl^- goes in front of Pb^2+
[Pb^2+] = 0.010 M [Cl^-] = 0.020 M
2 [Pb^2+] = [Cl^-] 2(0.010) = 0.020
Mass Balance: 2 [Pb^2+] = [Cl^-]
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Ex. CaCO3 sol'n
CaCO3 <--> Ca^2+ + CO3^2-
[Ca^2+] = [CO3^2-] + [HCO3^-] + [H2CO3]
CO3^2- + H2O <--> HCO3^- + OH^-
HCO3^- + H2O <--> H2CO3 + OH^-
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Ex. AgCl sol'n
AgCl(s) <--> Ag^+ + Cl^-
[Ag^+] = [Cl^-] (Mass balance)
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Class exercise:
[Ca^2+] = 0.020 DO THIS!!!!!!!
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Class exercise:
[Ba^2+] = 0.020 M [CH3COO^-] + [CH3COOH] = 0.040 M
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Mass balance
Never put [H+] , [OH-] , or [H2O] in the equation
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Ex. 0.034 M carbonic acid
H2CO3 <--> HCO3^- + H^+
HCO3^- <--> CO3^2- + H^+
[CO3^2-] + [HCO3^-] + [H2CO3] = 0.034 M
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Ex. 0.10 M Na(CH3 COO)
[Na^+] = 0.10 M
Ch3COO^- + H2O <--> Ch3Cooh + OH^-
[Ch3Coo^-] + [CH3COOH] = 0.10 M
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Ex. 0.10 M CaCl 2
[Ca^2+] = 0.10 M
[Cl^-] = 0.10 M * 2 = 0.20 M
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Mass balance
CO3^2- + H2O <--> HCO3^- + OH^- Kb1
HCO3^- + H2O <--> H2CO3 + OH^- Kb2
[CO3^2-] + [HCO3^-] + [H2CO3] = 1g
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X. Systematic Treatment of Aqueous Equilibria.
Start EXAM 2
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