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Analytical Chemistry, Study guide for exam 2

by: Thomas Salazar

Analytical Chemistry, Study guide for exam 2 2154

Marketplace > Virginia Polytechnic Institute and State University > Chemistry > 2154 > Analytical Chemistry Study guide for exam 2
Thomas Salazar
Virginia Tech
GPA 3.32

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Notes cover pH buffers, monoprotic and polyprotic acid systems, and titrations
Majors Analytical Chemistry
Dr. Amanda Morris
Study Guide
Analytical Chemistry, pH, buffers, titrations
50 ?




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This 4 page Study Guide was uploaded by Thomas Salazar on Sunday September 25, 2016. The Study Guide belongs to 2154 at Virginia Polytechnic Institute and State University taught by Dr. Amanda Morris in Fall 2016. Since its upload, it has received 56 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
Study Guide for Exam 2 Analytical Chemistry PH BUFFERS  Buffer: a chemical solution containing significant amounts of a weak acid/base pair that resists large changes in pH o Even when diluted o In the presence/addition of strong acid or base  Generally, buffers will work within +/- 1 pH unit of the pH you are trying to maintain, this is the buffering range o For example, if you are trying to maintain a solution around pH 4.5, a buffer of acetic acid/acetate solution has a pKa of 4.75, so it would be able to buffer solutions within a range of 3.75 – 5.75 pH FRACTIONAL COMPOSITION  F total [HA] + [A-] o F total the formal concentration of a solution; the concentration of the original substance you are adding from a stock solution o Since all acids/bases react to some equilibrium, the F will always be some fraction of the base form, and some fraction of the acid form  Making Buffers o Buffers can be made from:  a weak acid and a salt of its conjugate base  a weak acid and enough strong base to form some of its conjugate base  a weak base and enough strong acid to form some of its conjugate acid  Polyprotic Acids and Bases H 2 + H O2 HA + H O- 3 + K a1 HA + H O2 A + H O2- 3 + K a2 F = [H2A]+[HA ]+[A ]2- o Adding base to diprotic acid (has two dissociable protons) causes the first acidic proton to dissociate virtually completely before the 2 ndacidic proton begins to be neutralized.  The acid dissociation constant for the 1 acidic proton = K = a1 1 Study Guide for Exam 2 Analytical Chemistry nd  For the 2 acidic proton, it is Ka2 K 2 o Finding the Concentration of Each Species at a Given PH  Following normal equilibrium constant expressions:  [HA-] = [H 2] K /a1H O 3 +  [A ] = [HA-] K /a2H O 3 ; substituting in the 1 equation, you get [A ] 2- = [H 2] K Ka2 a1 O ] 3 + 2  Accounting for mass conservation (F ), we end up with an equation: total + 2 + + 2 F total [H2A] * {([H3O ] + [H O3] K + 1 K )/1 2 O ] 3 +  Given pH we can calculate [H O ], 3nd the given the initial value of F we can calculate [H A]2 and subsequently derive the concentrations of the other two species in solution. In diprotic systems if the pH of solution = pK 1 α H2A = αHA- = 0.5, where α represents the fractional ratio of species (ergo, half of the total concentration is from the H A f2rm, and half is from the HA form.) At pK , th1 concentration of [A ] is virtually 0. If pH of solution = pK , then α = α = 0.5 {The same pattern follows} 2 HA- A2- DIPROTIC ACIDS: behave as if K = K … a 1  Since K is 2enerally negligibly small, we assume for the rxn. of H A with wa2er, the K aust equals K a1.  Likewise for the rxn. of A with water, we treat it as if K = K b b1 = 1/K a2  For the intermediate/amphiprotic form: o Acts as both an acid and base o Must use Systematic Treatment of Equilibria to solve pH problems o To find pH of a solution to which you add the intermediate form: + [H ] = √(K 1 2 + K K1/ w + F1  A simplified version of the equation can be used if: o The term K F >2 K AND w o K <1 F pH = ½ (pK + 1K ) 2 DIPROTIC BUFFERS: act roughly the same as normal buffers…  Simply, there are now 2 Henderson-Hasselbach eqns: - 2- - o pH = pk + 1og ([HA ]/[H A]) 2 or pH = pk 2 log ([A ]/[HA ]) Study Guide for Exam 2 Analytical Chemistry o Use whichever you are given values for POLYPROTIC SYSTEMS: follows similar ideas…  For H A 3 K = Ka a1 o For H A 2 [H ] = √(K K F1 2K K / K1+ w 1  For HA : [H ] = √(K K F + K K / K + F 2 3 2 w 2 3-  For A : K =bK b1 = Kw/K a3 DETERMINING PRINCIPLE SPECIES:  In monoprotic systems: o When pH > pK  A ia the principle form o When pH < pK  HA as the principle form  Polyprotic systems: o When pH = pk a1 [H A2 = [HA ]; when pH = pk a2  [HA ] = [A ]2- FRACTIONAL COMPOSITION IN POLYPROTIC SYSTEMS:  α H2A = [H 2]/F = [H ] /[H ] + [H ]K + 1 K 1 2 - - + + 2 +  α HA = [HA ]/F = [H ]K /1H ] + [H ]K + 1 K 1 2 2- 2- + 2 +  α A = [A ]/F = K 1 /2H ] + [H ]K + K1K 1 2 ISOELECTRIC POINT and ISOIONIC POINT:  Isoelectric point: the pH of a solution where the overall charge of the solution of the acid is 0; ergo, the charged species of a polyprotic acid are at equal 2- concentrations. (e.g. [H A2 = [A ])  Isoionic pH: the pH of a solution when only the uncharged intermediate, zwitterionic form of an acid (HA) is dissolved into aqueous solution. Obviously it will dissociate into its other possible species, but the charged species concentrations will not be equal, and the solution does not necessarily have a net neutral charge. SOLUTIONS OF INTERMEDIATE SPECIES of DIPROTIC SYSTEMS + 2- - o Charge balance eqn.  [H ] + [H A] = [A2] + [OH ] o Mass Balance  F = [H A] + [H2 ] + [A ]- 2- o HA  H + A + 2- (deprotonation of intermediate) o HA + H O 2H A + OH 2 - (protonation of intermediate) TITRATION Study Guide for Exam 2 Analytical Chemistry 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 V 1 C1V 2 2 o Titration of a Weak Acid with a Strong Base o e.g.) HF + KOH  H O + K2 ????2 1. Before No base added, treat as a weak acid problem. K = a , x = [H ] = [A ] ????−???? 2. Before 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 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 a1+ log ([HA ]/[H A2). Calculate concentrations based on how much strong base added.  At ½ V e1H = pK , [H Aa1= [H2 ], and [A ] is virtually 0.  At V e1pH = ½ (pK + pK )a1all spa2ies are in intermediate form 2- -  Between Eq.pts. 1 and 2: pH = pK a2 + log ([A ]/[HA ]) 2- -  At ½ V e2H = pK , [A ]a2 [HA ], and [H A] is virt2ally 0.  At V e2all is in fully deprotonated form A . Treat as weak base problem, as if Kb= K b1.


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