5.111 Lecture Summary #24

24.1 5.111 Lecture Summary #24 Topics: Oxidation States, and Balancing Oxidation/Reduction Reactions (Read Section K, Chapter 12) FROM MONDAY: Titrations Curves for Weak acid/Strong base and for Weak base/Strong acid 14 V > Veq strong base in water 12 V = Veq (salt) conj. base of weak acid 10 pH equivalence equivalence point >7 point >7 S 8 A- Strong base A- A- V=Vhalf-eq special category of buffer 7 H A 6 buffering region buffering region 4 0 < V < Veq buffer half-eq point 2 0 H A Weak acid half-eq point 0 A- Start V=0 H A = A- A- H A A- H A weak acid in water H A H A H A H A 10 20 30 Volume of base added (mL) Figure by MIT OpenCourseWare. V=0 14 B 12 buffering Strong acid region buffering region 10 pH B B B 0 < V < Veq buffer H B+ B B B V=Vhalf eq 8 7 6 equivalence point pH <7 4 S equivalence Weak base point pH < 7 2 0 weak base in water V = Veq V > Veq 0 10 20 30 Volume of acid added (mL) Figure by MIT OpenCourseWare. special category of buffer H B+ H B+ = B B (salt) conj. acid of weak base H B+ H B+ H B+ strong acid in water H B+ 24.2 Example: Titration of weak acid with strong base 25.0 mL of 0.10 M HCOOH with 0.15 M NaOH (Ka = 1.77 x 10-4 for HCOOH) 1. Volume = 0 mL of NaOH added Before any NaOH is added, the problem is that of an ionization of a weak acid in water. HCOOH (aq) + H2O (l) initial molarity change in molarity equilibrium molarity H3O+ (aq) + HCO2- (aq) HCOOH (aq) 0.10 M -x 0.10-x H3O+ (aq) + HCO2- (aq) 0 0 +x +x__ x x Ka = 1.77 x 10-4 = (x)2/(0.10-x) ~= (x)2/0.10 x = 0.00421 (check 0.00421 is 4.2% of 0.10) okay pH = -log [0.00421] = 2.38 (to how many sig figs?) 2. 0 < V < Veq In this range, the acid has been partly ionized by the strong base (buffering region). Calculate the pH of the solution resulting from the addition of 5.0 mL of 0.15 M NaOH. Because OH- is a stronger base than HCO2-, it reacts almost completely with HCOOH. HCOOH (aq) + OH- (aq) → H2O (l) + HCO2- (aq) K>>1 Initial Moles For HCOOH, (25.0 x 10-3 L)(0.10M) = 2.5 x 10-3 moles For OH-, (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 HCOOH left 0.75 x 10-3 moles OH- produces _____________of HCO2Molarity 1.75 x 10-3 moles of HCOOH/ (0.0250 + 0.0050 L) = 0.0583 M HCOOH 0.75 x 10-3 moles of HCO2- / (0.0250 + 0.0050 L) = 0.0250 M HCO2Option 1 HCOOH H3O+ + HCO2- initial molarity 0.0583 0 0.0250 change in molarity -x +x +x equilibrium molarity 0.0583 -x +x 0.0250 + x 24.3 Ka = 1.77 x 10-4 = (0.0250 + x )(x) assume x is small ~= 0.0250x (0.0583-x) 0.0583 x = 4.13 x 10-4 Check assumption: 4.13 x 10-4 is 1.65% of 0.025 and is 0.7% of 0.0583 okay pH = -log [4.13 x 10-4] = 3.38 Option 2 pH ~= pKa - log ([HA]/[A-]) pH ~= 3.75 - log ([0.0583]/[0.0250]) = 3.75 - 0.368 = 3.38 check assumption: for a pH of 3.38, [H3O+] = 4.2 x 10-4 and that is <5% of 0.0583 and is <5% of 0.0250. Okay If the 5% assumption is not valid, than option 1 must be used and Ka = 1.77 x 10-4 = (0.0250 + x)(x)/(0.0583 - x) can not be simplified. Must solve by quadratic equation. Note: when the volume of NaOH added is between 0 and the equivalence volume Veq, the problems are similar to buffer problems. This region of the titration curve is called the "buffering region." Half-equivalence point When the volume of NaOH added is equal to half the equivalence volume, [HA] = [A-]. pH ~= pKa - log ([HA]/[A-]) pH ~=pKa -log (1) pH ~=pKa 3. V = Veq At the equivalence point, the amount of NaOH added is equal to the amount of HCOOH. The pH is not 7 as it is for a strong acid and a strong base. The pH is >7 when a weak acid is titrated with a strong base. The pH depends on the properties of the salt formed during the neutralization process. HCOOH and NaOH form NaHCO2 and H2O. Na+ has ___________ on pH and HCO2- is ___________. Thus at the equivalence point, the pH is >7. 24.4 Calculate the pH at the equivalence point Calculate total volume at equivalence point moles of HCOOH = 2.5 x 10-3 moles = moles of HCO2- formed = moles of OH- added 2.5 x 10-3 moles of OH- x 1L = 1.67 x 10-2 L of NaOH added 0.15 mol Total volume = 0.0250 L + 0.0167 L = 0.0417 L Molarity of HCO22.5 x 10-3 moles of HCO2-/ (0.0417 L) = 0.0600 M HCO2This is an ionization of weak base in water problem. HCO2- (aq) + H2O (l) HCOOH (aq) + OH- (aq) initial molarity change in molarity equilibrium molarity HCO2- (aq) 0.0600 -x 0.0600 -x HCOOH (aq) + OH- (aq) 0 0 +x +x +x +x You can take it from here. Simplify if x is small compared to 0.0600 M. Calculate x, which is equal to [OH- ] = 1.83 x 10-6 M. Then calculate pOH = 5.74. From pOH, calculate pH. pH = 8.26 (which is >7) 4. V > Veq Beyond the equivalence point, NaOH is added to the solution of the conj. base HCO2-. Since HCO2does not give rise to much OH- in solution (1.83 x 10-6 M), the pOH and pH are determined by the amount of excess NaOH added. This problem is similar to a strong acid/strong base problem. At 5.00 mL past the equivalence point: 0.00500 L x 0.15 M = 7.5 x 10-4 moles excess OH 7.5 x 10-4 moles OH- /(0.00500 L + 0.0250 L+ 0.0167 L) = 0.016 M OH pOH = -log [0.16] = 1.79 pH = 12.21 24.5 Today’s material OXIDATION/REDUCTION REACTIONS Guidelines for assigning oxidation numbers 1) In free elements, each atom has an oxidation number of zero. Example H2 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. Aluminum has an oxidation number of +3 in all its compounds. 3) The oxidation number of oxygen in most compounds is -2. However, in peroxides such as H2O2 and O2-2, 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 LiH, NaH, CaH2. 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). However, 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 NH4+ H is N is Sum is 7) Oxidation numbers do not have to be integers. For example, the oxidation number of oxygen in superoxide O2-1 is _________ Examples Li2O PCl5 HNO3 N2O Definitions Oxidation Reduction Oxidizing agent Reducing agent 24.6 Disproportionation Reaction A reactant element in one oxidation state is both oxidized and reduced. NaClO ⇒ NaClO3 + NaCl in basic solution Write the half reactions and determine the changes in oxidation state. Na+ is a spectator ion so: ClO- ⇒ ClO3- ClO- ⇒ Cl- Balancing Redox Reactions (Ch12.2) A. BALANCE IN ACIDIC SOLUTION Fe2+ + Cr2O72- ⇒ Cr3+ + Fe3+ (1) Write two unbalanced half reactions for oxidized and reduced species. Cr2O72- ⇒ Cr3+ Fe2+ ⇒ Fe3+ (2) Insert coefficients to make the number of atoms of all elements except oxygen and hydrogen equal on the two sides of each equation. Cr2O72- ⇒ Cr3+ Fe2+ ⇒ Fe3+ (3) Add H2O to balance oxygen Cr2O72- ⇒ 2Cr3+ Fe2+ ⇒ Fe3+ (4) Balance hydrogen with H+ Cr2O72- ⇒ Fe2+ ⇒ Fe3+ 2Cr3+ + 7H2O 24.7 (5) Balance the charge by inserting electrons 14H+ + Cr2O72- ⇒ 2Cr3+ + 7H2O Fe2+ ⇒ Fe3+ (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- + 14H+ + Cr2O72- ⇒ 2Cr3+ + 7H2O Fe2+ ⇒ Fe3+ + e (7) Add half reaction, make appropriate cancellations. 6e- + 14H+ + Cr2O72- + 6Fe2+ ⇒ 2Cr3+ + 7H2O + 6Fe3+ + 6e B. BALANCE IN BASIC SOLUTION (Book has a different approach. You can use either.) Fe2+ + Cr2O72- ⇒ Cr3+ + Fe3+ Follow steps (1-7) to get your answer for acidic solution: 14H+ + Cr2O72- + 6Fe2+ ⇒ 2Cr3+ + 7H2O + 6Fe3+ (8) Then "adjust pH" by adding OH- to both sides to neutralize H+. 14OH- + 14H+ + Cr2O72- + 6Fe2+ ⇒ 2Cr3+ + 7H2O + 6Fe3+ + 14OH OR 14H2O + Cr2O72- + 6Fe2+ ⇒ 2Cr3+ + 7H2O + 6Fe3+ + 14OH CANCEL 7 14H2O + Cr2O72- + 6Fe2+ ⇒ 2Cr3+ + 7H2O + 6Fe3+ + 14OH Thus: 7H2O + Cr2O72- + 6Fe2+ ⇒ 2Cr3+ + 6Fe3+ + 14OH