Week 5 notes
Week 5 notes 2154
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This 2 page Class Notes was uploaded by Thomas Salazar on Sunday September 25, 2016. The Class Notes belongs to 2154 at Virginia Polytechnic Institute and State University taught by Dr. Amanda Morris in Fall 2016. Since its upload, it has received 47 views. For similar materials see Majors Analytical Chemistry in Chemistry at Virginia Polytechnic Institute and State University.
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Date Created: 09/25/16
Note Set #3 – week 5 Analytical Chemistry SOLUTIONS OF INTERMEDIATE SPECIES of DIPROTIC SYSTEMS + 2- - o Charge balance eqn. [H ] + [H A] = [A 2 + [OH ] - 2- o Mass Balance F = [H A] + [HA2] + [A ] - + 2- o HA H + A (deprotonation of intermediate) o HA + H O 2H A + OH 2 - (protonation of intermediate) TITRATION o Titration of a Strong Acid with a Weak Base o e.g.) HI + KOH H O + 2I *goes to completion* 1. Before equivalence point: pH is dependent on excess, unreacted/un-neutralized H+ 2. At equivalence point: pH = 7.00; [H+] is equivalent to [OH-] 3. After equivalence point: pH is dependent on excess, unreacted/un-neutralized OH- Useful eqn.: C V1= 1 V 2 2 o Titration of a Weak Acid with a Strong Base o e.g.) HF + KOH H O + K2 ????2 + - 1. Before eq.pt: No base added, treat as a weak acid problem. K = a ????−???? , x = [H ] = [A ] 2. Before eq.pt: Some base added, pH = pK + log ([A a/[HA]). Moles HA = Moles HA – i Moles OH added. Moles A = Moles A + Moles OH added. Make sure to convert units back to molar concentrations (if you have to). 3. At eq.pt: All weak acid converted to weak base, treat as weak base problem. K = b ????2 K wK a , where F is the new concentration of [A ], and x= [OH ]. - ????−???? 4. Post-equivalence: pH dependent on excess, unreacted/un-neutralized OH- *When titrating Weak Base with Strong Acid, simply work from all weak base to begin with, and end with all of it converting to weak acid.* POLYPOTIC TITRATIONS Titration of Weak diprotic acid, with Strong Base o Points to recognize At beginning: no base added, treat H A as a2weak monoprotic acid if K = K a 1 Before equivalence pt. 1 (V ): pe1= pK + log a1HA ]/[H A]). Calc2late concentrations based on how much strong base added. Note Set #3 – week 5 Analytical Chemistry - 2- At ½ V e1 = pK , [H A] a1[HA 2, and [A ] is virtually 0. At V e1H = ½ (pK + pK ), a1l speca2s are in intermediate form 2- - Between Eq.pts. 1 and 2: pH = pK a2 + log ([A ]/[HA ]) 2- - At ½ V e2 = pK , [A ] =a2HA ], and [H A] is virtua2ly 0. At V all is in fully deprotonated form A . Treat as weak base problem, as e2 if Kb= K b1.
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