CHEM 1332 1st Edition Lecture 14 Outline of Last Lecture - Problem- Acids- Percent errorOutline of Current Lecture - Arrhenius acids and bases- Bronstel and Lowry - Conjugate pairs- Relationship with KCurrent Lecture- Acids according to Arrhenius in water the formula is −¿(aq)+¿(aq)+ A¿HA(aq)⇄ H¿-+¿ eqA−¿eqHAeqH¿Ka(HA)=¿- Strong acids have a Ka so big that we assume the equation is −¿(aq)+¿(aq)+A¿HA(aq)→ H¿- The strong acids are HCl(aq), HBr(aq), HI(aq), HN O3(aq), HCl O4(aq), H2S O4(aq)-+¿eqH¿ pH=-log+¿eqH¿- Base dissociates in −¿(aq)H2O →O H¿- Strong bases areLiOH(aq), NaOH(aq), KOH , RbOH(aq), CrOH(aq), Sr(OH)2(aq), BaCO H2(a q)- Weak bases are all the other bases- Bronstel and Lowry had a different way of thinking they said that acids were +¿H¿donors and bases were +¿H¿acceptors 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.-−¿(aq)HA(aq)+H2O(l)⇄ H3O(aq)+A¿-+¿e qA−¿eqHA eqH3O¿Ka(HA)=¿-+¿H3O¿is the same as +¿H¿-+¿+¿=− log H3O¿pH=−log H¿- Base −¿(aq)+¿(aq)+O H¿B(aq)+H2O(l)⇄ B H¿-+¿eqOH−¿ eqBeqBH¿Kb=¿- Percent error is also known as reaction ionization dissociation - Conjugate pairs +¿(aq)−¿(aq)+B H¿HA(aq)+B(aq)⇄ A¿-−¿(aq)HA(aq)∧ A¿ are conjugate pairs to HA(aq) is the acid and −¿(aq)A¿ is the base and to get to from the acid to base you lose +¿H¿ and to get from base to acid you gain-+¿(aq)B(aq)∧B H¿ are conjugate pairs and B is the base while +¿(aq)B H¿ is the acid and the same rules apply- Strong acids have very weak conjugate bases - Weak acids have strong conjugate bases-Ka(HA) is proportional to A¿−¿Kb¿¿- If HA is stronger than +¿H3O¿ than K is greater than 1- If HA is weaker than +¿H3O¿ than K is less than 1- What depends on the reaction- It is amphiprotic which means both and protonKey TermsAmphiprotic- having characteristics of both an acid and a base and capable of reacting as
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