CHEM 1332 1st Edition Lecture 16 Outline of Last Lecture - pH and pOH levels - Types of Acids- Types of BasesOutline of Current Lecture - Anions- Cations- Polyprotic AcidsCurrent Lecture- Anions are conjugate bases of acids- Strong bases −¿−¿C H3¿+¿, N H2¿−¿, H¿OH¿ hydrocarbon ion- Very weak acids −¿−¿ ,C lO4¿− ¿ , N O3¿−¿, I¿−¿, B r¿C l¿ these are either weak acids or neutral conjugates- An example is −¿(aq)+HF(aq)−¿(aq)+H2O ⇄O H¿F¿-Kb(F)=[OH]eq[HF]eq[F]eq-−¿A¿¿Ka(HA)× Kb ¿- Cations are conjugate of acids or bases +¿B ⇄ B H¿¿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.-+¿+N H3(aq)+¿(aq)+H2O ⇄ H3O¿N H4¿-Ka(N H4)× Kb(N H3)=Kw- Metal Cations +¿M¿¿-+¿(aq)+H2O ⇄ MM¿- Weak acids 2+¿2+¿ ,B a¿+¿ , S r¿+¿ , R b¿+¿ , K¿+¿, N a¿L i¿- Polyprotic Acids is HnA- A strong acid acts like −¿(aq)+¿(aq)+H3S O4¿H2S O4(aq)+H2O(l)→ H3O¿-2−¿(aq)+¿(aq)+S O4¿−¿(aq)+H2O(l)→ H3O¿H3S O4¿- Weak acid −¿(aq)Ka1+¿(aq)+H2Sb O4¿−¿(aq)+H2O(l)⇄ H3O¿H3Sb O4¿-2−¿(aq)Ka2+¿(aq)+HSb O4¿−¿(aq)+H2O(l)⇄ H3O¿H2Sb O4¿-3−¿(aq)Ka3+¿(aq)+Sb O4¿2−¿(aq)+H2O(l)⇄ H3O¿HSbO4¿- Ka1>>Ka2>>Ka3-Ka1=1 x 1 0−5 Ka2=1 x 10−9 Ka1=1 x 1 0−14 - Calculate pH of 0.1 M −¿H3Sb O4¿-−¿(aq)+¿(aq)+H2Sb O4¿−¿(aq)+H2O(l)⇄ H3O¿H3Sb O4¿- Create a SCE chart to solve the problem0.1 - 0 0-x - X X0.1-x - X X- Ka−¿(aq)=x20.1−xH3Sb O4¿-1 x 10−5=x20.1 You need to take into consideration that the Ka value is very small and calculate from there-x2=1 x 10−6-x=1 x 1 0−3-2− ¿(aq)+¿(aq)+HSb O4¿−¿(aq)+H2O(l)⇄ H3O¿H2Sb O4¿1 x 10−3-1 x 10−30-X - X X1 x 10−3-X -1 x 10−3+¿X X- Ka2=1 x 10−9=(1 x 1 0−3+x)(x )1 x 10−3−x-1 x 10−3+x =1 x 10−3-x=1 x 1 0−9Ka 2- pH=-log (+¿H3O¿=− log 1 x 10−3=3 you can get this from only the first reaction-2−¿HSbO4¿eq=Ka2 which is the second conjugate baseKey TermsPolyprotic acid- have more than one ionizable
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