5.111 Principles of Chemical Science

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1 MIT OpenCourseWare Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit:

2 Lecture Summary #24 Topics: Oxidation States, and Balancing Oxidation/Reduction Reactions (Read Section K, Chapter 12) FROM MONDY: Titrations Curves for Weak acid/strong base and for Weak base/strong acid V > V eq strong base in water p equivalence point >7 S Strong base V = V eq (salt) conj. base of weak acid V=V halfeq special category of buffer = 4 2 buffering region halfeq point Weak acid 0 < V < V Start V = 0 eq buffer weak acid in water Volume of base added (ml) Figure by MIT OpenCourseWare. V = 0 weak base in water 14 B B B B buffering region Strong acid 0 < V < V buffer eq B + B B B p equivalence point p <7 S Weak base V=V half eq V = V eq special category of buffer B + B + = B B (salt) conj. acid of weak base B + B + B + B + 2 V > V eq strong acid in water Volume of acid added (ml) Figure by MIT OpenCourseWare.

3 24.2 Example: Titration of weak acid with strong base 25.0 ml of 0.10 M COO with 0.15 M NaO (K a = 1.77 x 10 4 for COO) 1. Volume = 0 ml of NaO added Before any NaO is added, the problem is that of an ionization of a weak acid in water. COO (aq) + 2 O (l) 3 O + (aq) + CO 2 (aq) COO (aq) initial molarity 0.10 M 0 0 change in molarity x +x +x equilibrium molarity 0.10x x x K a = 1.77 x 10 4 = (x) 2 /(0.10x) ~= (x) 2 /0.10 x = (check is 4.2% of 0.10) okay p = log [ ] = 2.38 (to how many sig figs?) 3 O + (aq) + CO 2 (aq) 2. 0 < V < V eq In this range, the acid has been partly ionized by the strong base (buffering region). Calculate the p of the solution resulting from the addition of 5.0 ml of 0.15 M NaO. Because O is a stronger base than CO 2, it reacts almost completely with COO. COO (aq) + O (aq) 2 O (l) + CO 2 (aq) K>>1 Initial Moles For COO, (25.0 x 10 3 L)(0.10M) = 2.5 x 10 3 moles For O, (5.0 x 10 3 L)(0.15M) = 0.75 x 10 3 moles Moles after Reaction 2.5 x 10 3 moles 0.75 x 10 3 moles = 1.75 x 10 3 moles of COO left 0.75 x 10 3 moles O produces of CO 2 Molarity 1.75 x 10 3 moles of COO/ ( L) = M COO 0.75 x 10 3 moles of CO 2 / ( L) = M CO 2 Option 1 COO 3 O + + CO 2 initial molarity change in molarity x +x +x equilibrium molarity x +x x

4 K a = 1.77 x 10 4 = ( x )(x) assume x is small ~= x (0.0583x) x = 4.13 x 10 4 Check assumption: 4.13 x 10 4 is 1.65% of and is 0.7% of p = log [4.13 x 10 4 ] = okay Option 2 p ~= pk a log ([]/[ ]) p ~= 3.75 log ([0.0583]/[0.0250]) = = 3.38 check assumption: for a p of 3.38, [ 3 O + ] = 4.2 x 10 4 and that is <5% of and is <5% of Okay If the 5% assumption is not valid, than option 1 must be used and K a = 1.77 x 10 4 = ( x)(x)/( x) can not be simplified. Must solve by quadratic equation. Note: when the volume of NaO added is between 0 and the equivalence volume V eq, the problems are similar to buffer problems. This region of the titration curve is called the "buffering region." alfequivalence point When the volume of NaO added is equal to half the equivalence volume, [] = [ ]. p ~= pk a log ([]/[ ]) p ~=pka log (1) p ~=pka 3. V = V eq t the equivalence point, the amount of NaO added is equal to the amount of COO. The p is not 7 as it is for a strong acid and a strong base. The p is >7 when a weak acid is titrated with a strong base. The p depends on the properties of the salt formed during the neutralization process. COO and NaO form NaCO 2 and 2 O. Na + has on p and CO 2 is. Thus at the equivalence point, the p is >7.

5 Calculate the p at the equivalence point 24.4 Calculate total volume at equivalence point moles of COO = 2.5 x 10 3 moles = moles of CO 2 formed = moles of O added 2.5 x 10 3 moles of O x 1L = 1.67 x 10 2 L of NaO added 0.15 mol Total volume = L L = L Molarity of CO x 10 3 moles of CO 2 / ( L) = M CO 2 This is an ionization of weak base in water problem. CO 2 (aq) + 2 O (l) COO (aq) + O (aq) CO 2 (aq) COO (aq) + O (aq) initial molarity change in molarity x +x +x equilibrium molarity x +x + x You can take it from here. Simplify if x is small compared to M. Calculate x, which is equal to [O ] = 1.83 x 10 6 M. Then calculate po = From po, calculate p. p = 8.26 (which is >7) 4. V > V eq Beyond the equivalence point, NaO is added to the solution of the conj. base CO 2. Since CO 2 does not give rise to much O in solution (1.83 x 10 6 M), the po and p are determined by the amount of excess NaO added. This problem is similar to a strong acid/strong base problem. t 5.00 ml past the equivalence point: L x 0.15 M = 7.5 x 10 4 moles excess O 7.5 x 10 4 moles O /( L L L) = M O po = log [0.16] = 1.79 p = 12.21

6 24.5 Today s material OXIDTION/REDUCTION RECTIONS Guidelines for assigning oxidation numbers 1) In free elements, each atom has an oxidation number of zero. Example 2 2) For ions composed of only one atom the oxidation number is equal to the charge on the ion. Thus Li +1 has an oxidation number of +1. Group 1 and group 2 metals have oxidation numbers of +1 and +2, respectively. luminum has an oxidation number of +3 in all its compounds. 3) The oxidation number of oxygen in most compounds is 2. owever, in peroxides such as 2 O 2 and O 2 2, oxygen has an oxidation state of 1. 4) The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds, such as Li, Na, Ca 2. In these cases, its oxidation number is 1. 5) F has an oxidation number of 1 in all its compounds. Other halogens (Cl, Br, and I) have negative oxidation numbers when they occur as halide ions in compounds (Ex. NaCl). owever, when combined with oxygen (oxoacids), they have positive oxidation numbers (Ex. ClO ). 6) In a neutral molecule, the sum of the oxidation numbers of all the atoms must be zero. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion. For example N 4 + is N is Sum is 7) Oxidation numbers do not have to be integers. For example, the oxidation number of oxygen in superoxide O 2 1 is Examples Li 2 O PCl 5 NO 3 N 2 O Definitions Oxidation Reduction Oxidizing agent Reducing agent

7 Disproportionation Reaction reactant element in one oxidation state is both oxidized and reduced NaClO NaClO 3 + NaCl in basic solution Write the half reactions and determine the changes in oxidation state. Na + is a spectator ion so: ClO ClO 3 ClO Cl Balancing Redox Reactions (Ch12.2). BLNCE IN CIDIC SOLUTION Fe 2+ + Cr 2 O 7 2 Cr 3+ + Fe 3+ (1) Write two unbalanced half reactions for oxidized and reduced species. Cr 2 O 7 2 Cr 3+ (2) Insert coefficients to make the number of atoms of all elements except oxygen and hydrogen equal on the two sides of each equation. Cr 2 O 7 2 Cr 3+ (3) dd 2 O to balance oxygen Cr 2 O 7 2 2Cr 3+ (4) Balance hydrogen with + Cr 2 O 7 2 2Cr O

8 (5) Balance the charge by inserting electrons Cr 2 O 7 2 2Cr O (6) Multiply the half reactions so that the number of electrons given off in the oxidation equals the number of electrons accepted in the reduction. 6e Cr 2 O 7 2 2Cr O + e (7) dd half reaction, make appropriate cancellations. 6e Cr 2 O 7 + 6Fe 2+ 2Cr O + 6Fe e B. BLNCE IN BSIC SOLUTION (Book has a different approach. You can use either.) Fe Cr 2 O 7 Cr 3+ + Fe 3+ Follow steps (17) to get your answer for acidic solution: Cr 2 O 7 + 6Fe 2+ 2Cr O + 6Fe 3+ (8) Then "adjust p" by adding O to both sides to neutralize +. 14O Cr 2 O Fe 2+ 2Cr O + 6Fe O OR 14 2 O + Cr 2 O Fe 2+ 2Cr O + 6Fe O CNCEL O + Cr 2 O Fe 2+ 2Cr O + 6Fe O Thus: 7 2 O + Cr 2 O Fe 2+ 2Cr Fe O

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