Analytical Chemistry 2154, week 3 notes
Analytical Chemistry 2154, week 3 notes 2154
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This 2 page Class Notes was uploaded by Thomas Salazar on Sunday September 18, 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 6 views. For similar materials see Majors Analytical Chemistry in Chemistry at Virginia Polytechnic Institute and State University.
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Date Created: 09/18/16
Note Set #1 – week 3: September 7 – 14 , 2016 Analytical Chemistry SYSTEMATIC TREATMENT OF EQUILIBRIA Charge balance, solutions have an overall neutral charge, therefore the negative ions must equal the positive ions the sum of all the negative charges must equal the sum of all the positive charges o If [P] is the conc. of a cation, [N] is the conc. of an anion, z is the absolute value, or magnitude of the positive charge, and y is the absolute value or magnitude of the negative charge, then: z [P ] + 1 [1 ] … 2 y2[N ] + y 1N ]1 2 2 o The magnitude of an ion’s charge becomes its coefficient in the eqn. Mass balance, all species that exist in the final solution must come from the original substance added to the solution o activity is irrelevant in mass balance, since you are just counting the atoms of the species in solution o the quantity of a given molecule in all species that contain it in the solution, must be equal to the amount of that molecule in the original substance. + + - - + + - - o When dissolving solids with the equation: x a x a a x + a x The mass balance becomes: a [x ] = - [x ] +he conc. of the cation multiplied by the coefficient of the anion, equals the conc. of the anion multiplied by the coefficient of the cation o When the rxn. produces polyprotic acids/bases, or dissociates into several different ions, the mass balance must account for these For example, if you add lead sulfate (PbSO ) to s4lution, the sulfate 2- – dissociates into SO 4 , but will also form HSO 4 as well as H 2O 4 Therefore, the concentrations of these three species must all equal the original concentration of sulfate added to the solution. If the [SO 4 added = 0.5 M, then [SO ] + [H4O ] + [H SO4] = 0.52M 4 Systematic treatment of equilibrium Use: Determining concentrations [c] of unknown species in reactions (rxns), using equilibrium constant expression, charge balance and mass balance eqns. Steps: 1) Write out the balanced equations for all rxns. pertinent to the problem, or involved in the net rxn. 2) Write out the charge balance eqn. All charged species should be accounted for 3) Write out mass balance equations. There are two general types a. For a dissolving solid b. For Polyprotic acids/bases, or ions that react further in the solution (e.g. H 2O ) 4 4) Write equilibrium constant expression for each balanced eqn. written in step 1) 5) Determine how many eqns. you have [# = eqns. in steps 2+3+4] Note Set #1 – week 3: September 7 – 14 , 2016 Analytical Chemistry a. Determine how many unknowns you have (which species are you trying to find the concentrations of?) 6) Solve algebraically for unknowns: as long as your # of eqns. ≥ # of unknowns, you can solve for them. *TIP* Mass balance eqns. are usually a good place to start, try substituting them into equilibrium expressions that fit best. Try to get two eqns. that are both in terms of a single variable. Once you solve for one unknown the rest become simper to solve for. ACID AND BASES + - -14 [H ][OH ] = K =w1 x 10 For finding pH of solutions with acid/base added: A) If [acid or base] is greater than or equal to 1 x 10 M, then the auto-ionization of water into H+ and OH- is negligible and can be ignored in calculations B) If [acid or base] is less than 1x 10 M, then the pH is considered to be 7. This is because the dissociation of acid/base at this concentration is so small that the change in pH is negligible C) However, If: 1x 10 M < [acid or base] < 1 x 10 M , then you must use systematic treatment to solve for the pH WEAK ACID BASE EQUILIBRIA For weak acids, the dissociation into conj. Base and H is essentially even: When HA H + A , then the resultant [H ] more-or-less equals [A-] As a result, we can assume that a basic form to determining pH of weak acid solns. Is + 2 + as follows: K a [H ] / (F - [H ]), where F = conc. of weak acid added Like in ICE diagrams, you can sometimes ignore the [H ] that appears in the denominator of this “weak acid equation,” since the concentration is relatively small. However, to be as accurate as possible, you take it into account, and then rearrange the eqn. to solve the quadratic formula and solve for [H ] as x.