CHEM 102 1nd Edition Lecture 25Outline of Last Lecture I. Conjugate Base PairsII. SaltsIII. Acid-Base ReactionsIV. pH of SaltV. SummaryOutline of Current Lecture I. Ka pHII. Percent HydrolysisIII. pH of SaltIV. Polyprotic AcidV. BuffersVI. pH of 0.30 M HC2H3O2VII. Common-Ion EffectVIII. ShortcutCurrent LectureI. Ka pH- Ka = H 3 O+¿¿NH 4+¿¿¿[NH 3]¿¿ =(x) (x)(0.20−x)∗¿∗usethe 100 x rule , m ultiply Ka by 100 if the bottom doesnot change by 0.20 . then−x is not needed∗¿∗¿-H 30+¿¿Ka=5.6e-10=x2o .20; x=1.1e-5 M ;¿- Ph = -log [H3O+] = -log(1.1e-5) = 4.96 which is acidicII. Percent Hydrolysis- % Hydrolysis = amount hydrolyzedinitial amountx 100- Example: there was initially 0.20 M NH4NO and 1.1e-5 M reacted with water-1.1e-5 M0.20 Mx 100=0.0055% hydrolysisIII. pH of SaltThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- What is the pH of 0.10M NaCN solution?- NaCn is a basic salt because it’s components are Na- and CN+ and it’s anion has a conjugate of a weak acid; therefore the anion reacts as a weak base. CN- + H2O ↔ HCN + OH-CN- H2O HCN OH-Initial 0.10M 0 0Changes -x X XEquilibrium0.10 – x X X Kb = [HCN] [OH][CN]=(x) (x)(0.10−x);2.5e-5=x20.10; x=1.6e-3 [OH] = 1.6e-3 pOH=-log[OH]=-log(1.6e-3)=2.80 pH = 14 – pOH = 14 – 2.80 = 11.20; solution is basicIV. Polyprotic Acid- A polyprotic acid has multiple Ka’s- Example Phosphoric acid H3PO4 + H2O ↔ H3O+ H2PO4(3-) Ka1= 7.5e-3 (this is when the first H+ is lost) H2PO4- + H20 ↔ H30+ + HPO4(2-) Ka2 = 6.2e-8 HPO4(2-) + H2O ↔ H3O+ + PO4(3-) Ka3 = 3.6e-13- Ka1>Ka2>Ka3- If Ka1 = 10 x Ka2, you do not need to calculate beyond Ka2V. Buffers- Buffers control the pH to keep it neutral- Requires a huge amount of acid or base to change the pH- Buffers resist pH changeVI. pH of 0.30 M HC2H3O2 (aka Ac)- HAc ↔ H+ + Ac-HAc H+ Ac-Initial 0.30 0 0Change -x X XEquilibrium 0.30 – x X X- Ka = 1.8e-5 = (x)(x)/0.30 becauuse 0.30 – 0.0018 is approx. 0.30- X = 0.002323 = [H]- pH = -log(x) = 2.63 - **in pH only the hundredths and thousandths, etc. count as significant figures.VII. Common-Ion Effect- Consider a solution on Acetic acid (Ac = C2H3O2)- HAc + H2O ↔ H3O+ + Ac-- If acetate ion is added Cheautier says equilibrium will shift to the left- pH of 0.30 M HAc and 0.30M NaAc HAc ↔ H+ + Ac-HAc H+ Ac-Initial 0.30 O 0.30Change -x X XEquilibrium 0.30-x X 0.30 + x Ka = 1.8e-5 =x (o.30+ x)(0.30−x)=x (becausethe+x ∧−x cango away because of the 100 x rule∧the 0.30 will cancel out) X= 1.8e-5 = [H+] pH = -log[H+] = - log(1.8e-5) = 4.74VIII. Shortcut- pH = pka + log(salt conjugate/weak acid)- pOH – pKb + log(salt conjugate/weak
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