CHEM 170 1st Edition Lecture 27Outline of Last Lecture I. Acids and BasesII. Auto Protolysis of WaterIII. pH = -log[H+]IV. Strong vs Weak Acids and Bases V. Examples Outline of Current Lecture VI. examplesCurrent LectureChapter 16 - Acids and Basesnote that − ¿+¿+O H¿H2O ↔H¿ Kw=1 ×10−14 at 25°Cex. If there is 0.23 mol NaOH in 2.8 L of water, then what is the pH?−¿+¿+O H¿NaOH → N¿−¿+¿+O H¿NaOH +H2O → H2O+N¿, where NaOH is the base the reactant H2O is the acid, the product H2O is the conjugate acid and −¿O H¿ is the conjugate base−¿O H¿¿¿keeping in mind that −¿+¿+O H¿H2O ↔ H¿ and +¿H¿¿−¿O H¿Kw=1 ×10−14=¿These 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.+¿H¿¿−¿O H¿¿¿¿pH=−log(1.22× 10−13)=12.91ex. If the pH of orange juice is 2.85 then what is the concentration of hydroxide ions?+¿H¿102.85=1.4 × 103=¿−¿OH¿¿+¿H¿¿¿¿ex. −¿+¿+C N¿HCN ↔ H¿+¿H¿¿CN−¿¿¿¿❑¿Ka=6.17× 10−10¿= pKa=9.21ex. −¿+¿+O H¿NH3+H2O↔ NH4¿note that Kw=KaKb for conjugate pairsThe stronger the bond in the company. We also between two molecules the conjugates strength. Leveling Effect = groups of acids that all essentially disassociate 100% in water (solvent)−¿−¿+C l¿HCl→ H3O¿ex. 0.1 M HCN, where the pH is 5.2.−¿+¿+C N¿HCN ↔ H¿ 10−5.2=6.3 ×10−6MHCN+¿H¿−¿C N¿0.10 0 0-x +x +x0.1-x X x+¿H¿¿−¿C N¿¿¿Ka=¿x=4
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