Principles of Chemistry II © Vanden BoutExam Wednesday NightPlaceUTC 2.102A Last name A-KUTC 2.112A Last name L-ZTime7:30-9:00We will start right at 7:30We will end right at 9:00get there earlyPrinciples of Chemistry II © Vanden BoutMakeup Exam Sunday nightPlaceTBDTime6:30-8:00Anyone who would like to can take the makeup examYou cannot take bothPrinciples of Chemistry II © Vanden BoutWhen are you planning to take the examA. Wednesday nightB. Sunday nightPrinciples of Chemistry II © Vanden BoutConverting pH and pOHKw = [H+][OH-]log(Kw) = log([H+][OH-])log(Kw) = log[H+]+ log[OH-]log(10-14) = log[H+]+ log[OH-]-14 = -pH - pOH14 = pH + pOHPrinciples of Chemistry II © Vanden BoutFor the next exam Which of the following would be more helpfulA. More worksheetsB. Suggested back of the book problemsC. Suggested problems on eduspaceD. otherPrinciples of Chemistry II © Vanden BoutBuffer Both HA and A-HA(aq) H+(aq) + A-(aq)HA H+ A-ICE[HA]0-x[HA]0 -xO+x+xKa =[H+][A-][HA][A-]0+x[A-]0 + x(x)([A-]0+x)[HA]0 -x=really 10-7assuming x << C=(x)([A-]0)[HA]0Principles of Chemistry II © Vanden BoutpH in a buffer solution[H+][A-]0[HA]0Ka ≈we have approximated a small change[H+][A-]0[HA]0log(Ka) ≈log = log[H+] +log[A-]0[HA]0pKa = pH - log[A-]0[HA]0Principles of Chemistry II © Vanden BoutpKa = pH - log[A-]0[HA]0initial weak acidinitial conjugate base[A-]0=[HA]0equal acid/basepH = pKa-log[A-]0[HA]0=0[A-]0 < [HA]0more acidpH < pKa-log[A-]0[HA]0> 0[A-]0 > [HA]0more basepH > pKa-log[A-]0[HA]0< 0Principles of Chemistry II © Vanden BoutWhat is the pOH of a 0.01M solution of HClO4?" A." " 1 " B." " 2" C." " 7"" D." " 10" E." " 12[H+] = 10-2 pH = 2 pOH =12Principles of Chemistry II © Vanden BoutStrong Acids and Bases"Strong" means one thingThe substance dissociates 100% in waterStrong AcidHCl(aq) H+(aq) + Cl-(aq)Ka =[H+][Cl-][HCl]≈∞Strong ElectrolyteNaCl(s) Na+(aq) + Cl-(aq)Ksp = [Na+][Cl-] ≈∞Principles of Chemistry II © Vanden BoutWhat is the pH of a 10-10M solution of HCl?" A." " 2 " B." " 4" C." " 7"" D." " 10" E." " very slightly less than 7Principles of Chemistry II © Vanden BoutWhen do we get into problems with approximationsWhat approximations are we makingTypically that [H+]0 = 0 no H+ at the startnot a problem along as the concentration of acid or base is large enoughwhat is large enough big compared to 10-7Principles of Chemistry II © Vanden BoutWhen do we get into problems with approximationsWhat approximations are we makingThat the change is smallwhat is required for thisK should be small (weak acid, weak base)The initial concentration should be largeC-x is approximately Cthis is a comparison between C and xPrinciples of Chemistry II © Vanden BoutFor which of this will our approximations fail?" A." " 0.1 M solution of sodium acetate " B." " 1 M solution of HF " C." " 10-6 M solution of benzoic acid"" D." " 0.5 M solution of HCl" E." " 0.2 M solution of NaOHPrinciples of Chemistry II © Vanden BoutThe pKa of HF is 3.18. What is the pH of solution of100 mL of 0.1 M HF and 100 mL of a 0.2 M NaF?" A."" slightly less than 3.18 " B."" 3.18 " C."" slightly more than 3.18"Principles of Chemistry II © Vanden BoutpKa = pH - log[A-]0[HA]0initial weak acidinitial conjugate baseif the initial acid and base are similar in concentration than the pH is close to the pKaFor the pH to be 1 unit different than the pKa the difference in concentrations must be at least 10
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