Chem 222 #28 Review Dec 2, 2004

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1 Chem 222 #28 Review Dec 2, 2004

2 Announcement Please receive your quiz from the TAs. If you have any corrections about your quiz, labs, and notebook, please ask for corrections by 5pm of Dec 3. Final exam will be held on Dec 7 (Tue) from 6 pm (2 hours) at Lecture Center (LC) B1. Don t forget a calculator We will not return your final Your grades (final and overall) will be e- mailed to you in the morning of Dec 9 (Thu). Make sure TA knows your If you have any concerns about grading of your final, please visit your TA by 1 PM Dec 13 (Mon). Do not visit TA unless your overall grade is within 2 % of the cut-off line. We will not accept any requests about your grade.

3 Final Total 150 points (25 % of the overall grade) Multiple Choices 15 questions (10 3points + 5 4points = 50) 5 Long questions (20 5 points = 100 points) 3 Extra Multiple Choices? In the final, you should study all the chapters we covered, but Ch 15 & Ch 26 are not included Overall Grading A 90, B 80, C 65, D 50, E <50 I can write a letter for recommendation for students with A grade

4 Chem 222 Final Exam Hints 1. Sodium hypobromite (NaOBr) was dissolved in a buffer solution at ph The equation for the buffer is as follows: HOBr? H + + OBr - a. What is the ratio [OBr - ]/[HOBr]? pk a (HOBr)=8.63 b. A buffer solution was prepared by dissolving 5.67 g of NaOBr (FW ) and an appropriate mass of HOBr (FW ) in 1.50 L. What mass of HOBr was needed to adjust the ph to 6.75? c. What will the new ph be if ml of M NaOH is added to the solution?

5 *Refer to Chapter 10 Buffers *Henderson-Hasselbach Equation for Buffers ph = pk a + log([a - ]/[HA]) (prepared from weak acid/conjugate base) ph = pk a + log([b]/[bh + ]) (prepared from weak base/conjugate acid) *Concentrations of Species in Buffer Molarity ([X]) = moles/liter [X] = (mass X added) (formula weight of X)(volume of solution) Note: Volume will cancel in the numerator and denominator of log term in H-H equation.

6 *Addition of Strong Acid or Base to Buffer Notes: *Strong acid will react with weak base, and strong base will react with weak acid completely. *Set up initial/final moles table to determine moles (and concentrations if necessary) of species after addition. *Use H-H to determine new ph. B + H +? BH + (strong acid + weak base) HA + OH -? A - + H 2 O (strong base + weak acid) HA OH - A - Initial moles (massha)* (FW HA) (voloh - )* ([OH - ]) (mass A - )* (FW A - ) Change -mol OH - -mol OH - +mol OH - Final moles molha-moloh - 0 mola - +moloh -

7 Koy Presentation 2) A ml of M diprotic acid H 2 A (pka 1 = 3.71, pka 2 = 8.13) is titrated with with M NaOH a) What is the ph of initial of H 2 A? H 2 A HA - + H X X X K a = 6.76 * 10-4 = X 2 / X, X = 4.34 * 10-3 [H + ]= 4.34 * 10-3 ph = 2.36 b) What is ph at first ½ equivalent point and first equivalent point? ph = pka 1 = 3.71 at first ½ equivalent point ph = ½(pKa 1 + pka 2 ) = ½( ) = 5.92 c) What is ph at second ½ equivalent point and second equivalent point? ph = pka 2 = 8.13 at second ½ equivalent point mol of H 2 A = * = 4.4 * * 10-3 mol H 2 A * 2 mol H + * 1 mol OH - 1 mol H 2 A 1 mol H + = 8.8 * 10-3 mol OH - required 8.8 * 10-3 mol OH - = M OH - * V (OH - ) V (OH - ) = 8.8 * 10-2 L Total V = =.176 L

8 4.4 * 10-3 mol H 2 A * 1 mol A 2- = 4.4 * 10-3 mol A 2-1 mol H 2 A [A 2- ] = 4.4 * 10-3 mol A 2- = M A L A 2- + H 2 O HA - + OH X X X K b = K w /K a2 = 1 * /7.41 * 10-9 = 1.35 * 10-6 = X 2 / X, X = 1.84 * 10-4 [OH - ] = 1.84 * 10-4 poh = 3.74 ph = = d) Draw schematic of diprotic system titration ) 3) A dibasic compound D has pkb 1 = 4.45 and pkb 2 = 8.10 Find the fractions in the form DH 2+ 2, DH + and D at ph 7.00 Hint: α H 2 A = [H + ] 2 / { H + ] 2 + H + 2+ K 1 + K 1 K 2 } α DH 2 α HA - = K 1 [H + ] / { H + ] 2 + H + K 1 + K 1 K 2 } α DH + α A 2- = K 1 K 2 / { H + ] 2 + H + K 1 + K 1 K 2 } α D

9 K 1 = K w /K b2 K 2 = K w /K b1 4) What is a correct answer about this diagram (below)? ph More acid More basic predominant H 3 A form H 2 A - HA 2- A [H 3 A] = 4 [H 2 A] = [A 3- ] ph = 5 ph = 6 a) 1 = pk 3 and 3 = pk 1 b) 2 = pk 2 and 5 = ½(pK 1 + pk 2 ) c) 3 = pk 2 and 6 = ½(pK 1 + pk 2 ) d) 4 = [A 2- ] and 6 = ½(pK 2 + pk 3 ) 6) What is a wrong answer? a) At isoionic point: [H + ] = [( K 1 K 2 F + K 1 K w ) / K 1 + F] b) At isoisoelectric point: ph = ½(pK 1 + pk 2 ) c) Intermediate form is an amphiprotic form d) Intermediate form has ph = 0

10 Hint from Medhat Chapter 18 & 21 You should know and study Beer s Law. Know the units of the variable Example the units of concentration is Molarity(M). Significant figures are important. You should know the Heisenberg uncertainty principle, how to use it. You should review section 18-1 The properties of light.

11 Make sure you know when to use the dilution factors and when not to. You should understand and solve the midterm exam and also the two sample exams sample exam for the midterm and sample exam for the final. Study the notes from the lecture they seem to help a lot. Good luck!

13 Long Questions 1) A solution of Ni 2+ is prepared by placing 200mL of unknown M Ni 2+ into 20.00L of distilled water L is removed and placed into a 2.00L volumetric flask and filled to the mark with water. Five ml of the original L solution is placed in a 250.0mL volumetric flask and filled to the mark with water. Then 100.0mL from both all three solutions is placed in a 1.000L volumetric flask and filled to the mark with water. The final solution is titrated with 100.0mL of EDTA solution. The EDTA solution was prepared by dissolving 15g of EDTA (MW=372.24) in 1.000L of H 2 O. You may assume that one mole of EDTA reacts with one mole of Ni 2+. a) Calculate the Molarity of the EDTA solution. b) Write down all of the dilution factors in chronological order. c) Calculate the concentration of Ni 2+ in the original solution using the titration data and dilution factors.

14 Here are Some Tips and Strategies Draw a picture. Label your picture. You may work forwards or backwards. Follow the Moles!!! Remember, the final answer is in Molarity, which is Moles divided by Liters. Since you are given the volume of the final container, you only need to find the absolute number of moles that go into that container.

15 Start by assigning the unknown. Which is X Molarity for this example. [X M Ni 2+ ]. In each step calculate how many moles are transferred from one solution to the next. Moles Transferred = V removed * Concentration V removed C 1 =V final C 2 C 2 =(V removed /V final )C 1 The ratio (V removed /V final ) is the dilution factor.

16 Long Question Q1) We titrate ml of solution containing 2.00 mm Fe 2+ and 1.00 M H 2 SO 4 with V ml of mm MnO 4-. The net reaction is given by MnO Fe H + Mn Fe 3+ +4H 2 O (A) (a) How much volume of the MnO 4- solution (V e ) is required to reach the equivalence point? (b) The cell potential is given by E =1.507 ( /5)Log{[Mn 2+ ]/([MnO 4- ][H + ] 8 )} E = Log{[Fe 2+ ]/[Fe 3+ ]} Calculate E at the equivalence point. (c) Calculate E at V = 5.0 ml (d) Calculate E at V = 30.0 ml (you can neglect H + produced from (A)) Study well Ch. 16 P352. Study Dilution Factor

17 C 2 =(V 1 /V 2 )C L of X M Ni 2+ C 1 =X M Ni2+ V 1 =0.200L 20.0L H 2 O C 2 =??? V 2 =20.2L C 2 =(0.200L/20.2L) X M Ni 2+ DF1 =

18 2L C 3 =(V 2 /V 3 )C 2 C 2 = X M Ni L V 2 =1.000L V 3 =2.00L DF2 = C 3 = C 4 =(V 5 /V 4 )C 2 250mL

19 Dilution factor C 1 C 2 C 3 C4 f 12 f 23 f 34 C 1 V 1 = C 2 V 2 C 2 = (V 1 /V 2 )C 1 f 12 Mixing Two Solution C A C B C C (a) C C = C A + C B (b)c C = C A +C B (c) C C =(C A +C B )/2 (d) C C =(C A V A +C B V B )/(V A +V B ) Mixing Three Solution C A C B C D C C

20 2L 1L 20.2L 250mL

21 Check p352 Q1) We titrate ml of solution containing 2.00 mm Fe 2+ and 1.00 M H 2 SO 4 with V ml of mm MnO 4-. The net reaction is given by MnO Fe H + Mn Fe H 2 O (A) (a) How much volume of the MnO 4- solution (V e ) is required to reach the equivalence point? (b) The cell potential is given by E = ( /5)Log{[Mn 2+ ]/([MnO 4 -][H + ] 8 )} E = Log{[Fe 2+ ]/[Fe 3+ ]} Calculate E at the equivalence point. (c) Calculate E at V = 5.0 ml (d) Calculate E at V = 30.0 ml (you can neglect H + produced from (A)) You do not need to consider a dilution factor to calculate the ratio [Fe2+]/[Fe3+] or [Mn]/[MnO4-], but calculation of [H+] requires a dilution factor You don t have to calculate concentrations to calculate [Fe2+]/[Fe3+] and [Mn]/[MnO4-]

22 Titration of 10.0 ml of M Base (pk b1 =4.00, pk b2 = 9.00) with M HCl B/BH + BH + /BH 2 + B + H 2 O BH + + OH - F x x x x 2 /(F-x) = K b1 ~ (pk 1 + pk 2 )/2 E BH 2+ BH + + H + x 2 /(F-x) = K a1 = K w /K b2 F x x x

23 Modified Question from Quiz 5 We titrate 10.0 ml of M dibasic base (B) with M HCl. pk b1 = 4.0 and pk b2 = 9.0. Suppose V a ml of the HCl solution is titrated. (pk a1 = pk a2 = ) (a) Choose the value closest to the ph obtained when V a = 10.0 ml. (i) 5.0 (ii) 5.5 (iii) 6.0 (iv) 6.5 (v) 7.0 (vi) 7.5 (b) Choose the closest value to the ph when V a = 2.0 ml. (i) 3.0 (ii) 5.0 (iii) 7.0 (iv) 9.0 (v)11.0

24 What is it?

25 11-2 Diprotic Buffers H 2 A HA - A 2- You can use either or both of the equations ph = pk 1 + Log[HA - ]/[H 2 A] ph = pk 2 + Log[A 2- ]/[HA - ] Ex. Find the ph of a solution prepared by dissolving 1.0 mmol of KHP and 2.0 mmol of Na 2 P in a 1L of H 2 O. ph = pk 2 + Log{[P 2- ]/[HP - ]} [P 2- ]/[HP] =

Chem 222 #29 Review Apr 28, 2005

Chem 222 #29 Review Apr 28, 2005 Announcement Please meet me after the class if you have any conflicts with the final exam schedule. You can expect similar questions with Quiz 6 in the final exam. If you