Chem 222

#28 ReviewDec 2, 2004

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 e-mail • 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.

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 GradingA ≥ 90, B ≥ 80, C ≥ 65, D ≥ 50, E <50I can write a letter for recommendation

for students with A grade

Ch e m 222 Fin al Exam Hin ts

1. Sodium hypobromite (NaOBr) was dissolved in a buffer solut ion a t pH 6.75. The equa t ion for the buffer is as follows:

HOBr ? H + + OBr -

a . What is the ra t io [OBr -]/[HOBr]?

pKa(HOBr)=8.63 b. A buffer solut ion was prepared by

dissolving 5.67 g of NaOBr (FW 118.893) and an appropr ia te mass of HOBr (FW 96.911) 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 10.00 mL

of 0.0500 M NaOH is added to the solut ion?

*Re fe r to Ch apte r 10…Bu ffe rs

*He nde rson-Hasse lbach Equation for Bu ffe rs

pH = pKa + log([A-]/[HA])

(prepared from weak acid/conjugate base)

pH = pKa + log([B]/[BH+]) (prepared from weak base/conjuga te acid)

*Conce ntrations of Spe c ie s in Bu ffe r

Molar ity ([X]) = moles/liter

[X] = (mass X added) --------------------------------------------------(formula weight of X)(volume of solu t ion)

Note: Volume will cancel in the numera tor and denominator of log term in H-H equa t ion .

*Addition of Strong Acid or Base to Bu ffe r

Notes: *St rong acid will react with weak base, and st rong base will react with weak acid completely. *Set up in it ia l/fina l moles table to determine moles (and concent ra t ions if necessary) of species a fter addit ion . *Use H-H to determine new pH.

B + H + ? BH +

(st rong acid + weak base)

HA + OH - ? A- + H 2O (st rong base + weak acid)

HA OH - A-

In it ia l moles (massHA)* (FW HA)

(volOH -)*([OH -])

(mass A-)* (FW A-)

Change -mol OH - -mol OH - +mol OH - F ina l moles molHA-molOH - 0 molA-+molOH -

2) A 88.00 ml of 0.050 M diprotic acid H2A (pKa1 = 3.71, pKa2 = 8.13) is titrated with with 0.100 M NaOH a) What is the pH of initial of H2A? H2A HA- + H+

0.050 - X X X Ka = 6.76 * 10-4 = X2 / 0.050 – 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 = pKa1 = 3.71 at first ½ equivalent point pH = ½(pKa1 + pKa2) = ½(3.71 +8.13) = 5.92 c) What is pH at second ½ equivalent point and second equivalent point? pH = pKa2 = 8.13 at second ½ equivalent point mol of H2A = 0.050 * 0.088 = 4.4 * 10-3 4.4 * 10-3 mol H2A * 2 mol H+ * 1 mol OH-

1 mol H2A 1 mol H+

= 8.8 * 10-3 mol OH- required 8.8 * 10-3 mol OH- = 0.100 M OH- * V (OH-) V (OH-) = 8.8 * 10-2 L Total V = 0.088 + 0.088 = .176 L

Koy Presentation

4.4 * 10-3 mol H2A * 1 mol A2- = 4.4 * 10-3 mol A2- 1 mol H2A [A2-] = 4.4 * 10-3 mol A2- = 0.025 M A2- 0.176 L A2- + H2O HA- + OH- 0.025 - X X X Kb = Kw /Ka2 = 1 * 10-14 /7.41 * 10-9 = 1.35 * 10-6 = X2 / 0.025 – X , X = 1.84 * 10-4

[OH-] = 1.84 * 10-4 pOH = 3.74

pH = 14 – 3.74 = 10.26

d) Draw schematic of diprotic system titration

) 3) A dibasic compound D has pKb1 = 4.45 and pKb2 = 8.10 Find the fractions in the form DH2

2+, DH+ and D at pH 7.00 Hint: α H2A = [H+] 2 / { H+] 2 + H+K1 + K1K2} α DH2

2+ α HA- = K1 [H+] / { H+] 2 + H+K1 + K1K2} α DH+

α A2- = K1K2 / { H+] 2 + H+K1 + K1K2} α D

0

2

4

6

8

10

12

0 0.02 0.04 0.06 0.08 0.1

K1 = Kw /Kb2

K2 = Kw /Kb1 4) What is a correct answer about this diagram (below)? pH More acid More basic predominant form 1 2 3 [H3A] = 4 [H2A] = [A3- ] pH = 5 pH = 6

a) 1 = pK3 and 3 = pK1 b) 2 = pK2 and 5 = ½(pK1 + pK2) c) 3 = pK2 and 6 = ½(pK1 + pK2) d) 4 = [A2- ] and 6 = ½(pK2 + pK3)

6) What is a wrong answer? a) At isoionic point: [H+] = √[( K1K2F + K1Kw) / K1 + F] b) At isoisoelectric point: pH = ½(pK1 + pK2) c) Intermediate form is an amphiprotic form d) Intermediate form has pH = 0

H3A H2A- HA2- A3-

• 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”.

Hint from Medhat

• 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!

Long Questions1) A solution of Ni2+ is prepared by placing

200mL of unknown M Ni2+ into 20.00L of distilled water.

1.000L is removed and placed into a 2.00L volumetric flask and filled to the mark with water. Five mL of the original 20.00 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 H2O. You may assume that one mole of EDTA reacts with one mole of Ni2+.

a) Calculate the Molarity of the EDTA solution.

b) Write down all of the dilution factors in chronological order.

c) Calculate the concentration of Ni2+ in the original solution using the titration data

and dilution factors.

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.

Start by assigning the unknown. Which is X Molarity for this example. [X M Ni2+].

In each step calculate how many moles are transferred from one solution to the next.

Moles Transferred = Vremoved * Concentration

VremovedC1=VfinalC2

C2=(Vremoved/Vfinal)C1

The ratio (Vremoved/Vfinal) is the dilution factor.

Long Question Q1) We titrate 500.0 mL of solution containing 2.00 mM Fe2+ and 1.00 M H2SO4with V ml of 10.00 mM MnO4- . The net reaction is given by

MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3++4H2O

(A)(a) How much volume of the MnO4

- solution (Ve) is required to reach the equivalence point?

(b) The cell potential is given byE =1.507– (0.05916/5)Log{[Mn2+]/([MnO4

-][H+]8)} - 0.241

E = 0.680 – 0.05916 Log{[Fe2+]/[Fe3+]} -0.241 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

20.0L H2O

0.200L of X M Ni2+

C1=X M Ni2+V1=0.200L

C2=???V2=20.2L

C2=(V1/V2)C1

C2=(0.200L/20.2L) X M Ni2+

DF1 = 0.00990

20.2L

2L

250mL

C3=(V2/V3)C2

C4=(V5/V4)C2

C2 = 0.00990 X M Ni2+

V2=1.000LV3=2.00LDF2 = 0.500C3 = 0.00495

Dilution factor• C1 C2 C3 C4

f12 f23 f34

C1V1 = C2V2 C2 = (V1/V2)C1↑ f12

Mixing Two Solution• CA• CB CC

(a) CC = CA+ CB (b)CC = CA+CB(c) CC =(CA+CB)/2 (d) CC =(CAVA+CBVB)/(VA+VB)

Mixing Three Solution• CA• CB CD• CC

→→

→

1L

20.2L

2L

250mL

Q1) We titrate 500.0 mL of solution containing 2.00 mM Fe2+ and 1.00 M H2SO4with V ml of 10.00 mM MnO4- . The net reaction is given by

MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O

(A)(a) How much volume of the MnO4

- solution (Ve) is required to reach the equivalence point?

(b) The cell potential is given byE = 1.507 – (0.05916/5)Log{[Mn2+]/([MnO4-][H+]8)}

-0.241E = 0.680 – 0.05916 Log{[Fe2+]/[Fe3+]} -0.241

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-]

Check p352

Titration of 10.0 mL of 0.100 M Base (pKb1 =4.00, pKb2 = 9.00) with 0.100 M

HCl

B + H2O BH+ + OH- x2/(F-x) = Kb1

B/BH+ BH+/BH2+

E BH2+ BH+ + H+ x2/(F-x) = Ka1 = Kw/Kb2

F –x x x

F –x x x

~ (pK1 + pK2)/2

Modified Question from Quiz 5

• We titrate 10.0 mL of 0.100 M dibasic base (B) with 0.100 M HCl. pKb1 = 4.0 and pKb2 = 9.0. Suppose Va mL of the HClsolution is titrated.

(pKa1 = pKa2 = )

(a) Choose the value closest to the pH obtained when Va = 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 Va = 2.0 mL.

(i) 3.0 (ii) 5.0 (iii) 7.0 (iv) 9.0 (v)11.0

What is it?

11-2 Diprotic Buffers

You can use either or both of the equations

pH = pK1 + Log[HA-]/[H2A]pH = pK2 + Log[A2-]/[HA-]

Ex. Find the pH of a solution prepared by dissolving 1.0 mmol of KHP and 2.0 mmolof Na2P in a 1L of H2O.

pH = pK2 + Log{[P2-]/[HP-]}

[P2-]/[HP] =

A2-HA-H2A