Principles of Chemistry II © Vanden BoutpHLog scale. Useful when dealing with very small or very large number (big ranges of numbers)every "pH" unit is 10x larger or smaller [H+]pH = -log[H+]pH= 7[H+] =10-7pH= 2[H+] =10-2pH= 13[H+] =10-13Principles of Chemistry II © Vanden Boutlinear [H+]1M0 M10-1M pH =110-2 M pH =2pH 3-14log [H+]1M10-110-210-310-410-510-610-610-610-710-8pH=8 pH=0Principles 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 BoutStrong AcidsHCl HydrochloricHBr HydrobromicHI HydroiodicHClO4 PerchloricHClO3 ChloricH2SO4 SulfuricHNO3 NitricAll Dissociate 100%Principles of Chemistry II © Vanden BoutWhat is the pH of a 0.1 M solution of Nitric AcidHNO3(aq) H+(aq) + NO3-(aq)0.1 M acid makes a solution with [H+] = 0.1MpH =-log(0.1) = 1Principles of Chemistry II © Vanden BoutWhat is the pH of a 0.5M solution of HBr?# A.# # 0.5 # B.# # 1# C.# # 0.3## D.# # 0# E.# # 12[H+] =0.5 0<pH<1Principles of Chemistry II © Vanden BoutWe can ignore the conjugate base of a strong acidHCl(aq) H+(aq) + Cl-(aq)Ka =[H+][Cl-][HCl]≈∞equilibrium constant is so large,even if we add Cl- the shift back toHCl will be negligiblePrinciples of Chemistry II © Vanden BoutFor this reaction which has a higher entropy?H2O(l) H+(aq) + OH-(aq)# A.# # the products # B.# # the reactants# C.# # they are the samePrinciples of Chemistry II © Vanden BoutFor this reaction which has a lower enthalpy?H2O(l) H+(aq) + OH-(aq)# A.# # the products # B.# # the reactants# C.# # they are the samePrinciples of Chemistry II © Vanden BoutFor this reaction which has a lower free energy?H2O(l) H+(aq) + OH-(aq)# A.# # the products # B.# # the reactants# C.# # they are the samePrinciples of Chemistry II © Vanden BoutLiquid Water will spontaneously dissociate to a small extentH2O(l) H+(aq) + OH-(aq)K = [H+][OH-] 1Kw =[H+][OH-] = 10-14Principles of Chemistry II © Vanden BoutPure WaterH+ OH-ICEO+x+xO+x+xKw = 10-14 = [H+][OH-] = (x)(x)x = 10-7 [H+]=[OH-]=10-7Principles of Chemistry II © Vanden BoutpH of pure water at 25°Cx = 10-7 [H+]=[OH-]=10-7 pH = -log[H+]=-log(10-7) = 7Neutral[H+]=[OH-]at 25°CpH = 7pOH = 7Acidic[H+]>[OH-]at 25°CpH < 7pOH > 7[H+]<[OH-]at 25°CpH > 7pOH > 7BasicPrinciples of Chemistry II © Vanden BoutH2O(l) H+(aq) + OH-(aq)This reaction is endothermic.Given that information what do you think the pH is for pure water at 60°C?# A.# # 6.5 # B.# # 7# C.# # 7.5#Principles of Chemistry II © Vanden BoutIf pure water has a pH = 6.5 at 60°C is it Acidic?# A.# # Yes# B.# # No[H+]=[OH-] its neutral Kw = 9x10-14Principles of Chemistry II © Vanden BoutLet's look at three possible solutionsWeak AcidConjugate Base of the Weak AcidWeak Acid + Conjugate BasePrinciples of Chemistry II © Vanden BoutWeak AcidHA(aq) H+(aq) + A-(aq)HA H+ A-ICEC-xC-xO+x+xKa =[H+][A-][HA]O+x+x(x)(x)C-x=x ~ √KaCreally 10-7assuming x << CPrinciples of Chemistry II © Vanden BoutA-(aq) +H2O(l) HA(aq) + OH-(aq)Weak BaseKb =[HA][OH-][A-]identical result as before (same assumptions)[OH-] = √KbCPrinciples of Chemistry II © Vanden BoutBuffer Both HA and A-HA(aq) H+(aq) + A-(aq)HA H+ A-ICECHA-xCHA-xO+x+xKa =[H+][A-][HA]CA-+xCA- + x(x)(CA-+x)CHA-x=really 10-7assuming x <<
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