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Chem 126 Common Exam 2 March 21, 2016

1

last name f irst name (Middle)

Print Name:___________________│________________│____|

Instructor: Circle one: Butherus Gilbert Krishtal

Section #:___________Student #: _________

DIRECTIONS

(PUT AWAY YOUR CELL PHONE. IF YOU ARE CAUGHT WITH A CELL PHONE OR PDA OUTSIDE DURING THE EXAM, YOU WILL BE EJECTED FROM THE EXAM IMMEDIATELY AND

GIVEN A 0)

1. First fill in the information at the top of this paper. Print your last name as it appears on the class roster, then print your first name and middle initial. Then read and sign the honor code statement below. Finally fill in the requested information on the SCAN-TRON ANSWER SHEET according to the following scheme:

NAME: Print LAST name, then first (print clearly) SUBJECT: Course and section DATE: 03/21/16 PERIOD: Your Student Number 2. The last page of this exam is a periodic table. You may carefully remove this last page, if you choose, but do not open the staple.

3. This exam consists of two parts, A and B. PART A has 19 multiple choice questions worth four points each for a total of 76 points. Each question has only one correct answer, and no credit is lost if you choose a wrong answer. Therefore, leave no question unanswered, even if you simply must guess.

4. PART B consists of open ended problems worth 24 points. Show all work, including how units cancel to give the final answer, along with appropriate sig. figs. Write your final answers in the spaces provided. Reporting the correct answer without work will result in loss of credit.

HONOR CODE STATEMENT: On my honor I pledge that I have not violated the provisions of the NJIT student honor code.

Signature: —————————

___________ DO NOT WRITE BELOW THIS LINE _____________________

PART A: ______________ (76)

PART B: #1 ______________ (12)

#2 ______________ (12)

TOTAL SCORE:

Chem 126 Common Exam 2 March 21, 2016

2

Part A: 19 questions worth 4 points each for a total of 76 points.

1. A solution contains 5.5 × 10-5 M OH⁻ at 25°C. Calculate the concentration of H3O⁺ and identify the solution as acidic, basic or neutral.

A) 9.2 × 10-1 M, basic B) 1.8 × 10-10 M, basic C) 5.5 × 10-10 M, neutral D) 1.8 × 10-10 M, acidic E) 9.2 × 10-1 M, acidic Ans: B

1410

5

1.0 10[ ] 1.8 10 ,5.5 10

H basic�

� ��

u u

u

2. 40.0 mL of 0.20M HCl is added to 100.0 mL of 0.060M HNO3 What is the pH?

A) 0.70 B) 1.24 C) 1.37 D) 1.00 E) 2.74

Ans: D ( both strong acids, add moles and divide by total volume) (0.100 0.06 ) (0.040 0.20 )[ ] 0.10

0.14log 0.10 1.00

L M L MH ML

pH

� u � u

�

3. At a certain temperature, the pH of water is 6.63. Given that the autoionization of water is an endothermic reaction, choose the correct value of Kw and estimate the approximate temperature of this measurement.

A) 1.0× 10-14, 30 °C B) 2.34 × 10-14, 20 °C C) 5.50 × 10-14 , 60 °C D) 2.13 × 10-14, 40 °C E) There is not enough information to determine this.

Ans: C 6.63 7

7 2 7

[ ] [ ]10 2.34 10(2.34 10 ) 5.50 10w

H OHK

� � � �

� �

u

u u

Chem 126 Common Exam 2 March 21, 2016

3

3. An aqueous solution of 0.1M HC2H3O2 is added to a 0.1M NaOH solution. Which equation below describes the reaction accurately?

A) H+(aq) + OH-(aq) ÆH2O (l) B) C2H3O2

- (aq) + Na+(aq) Æ NaC2H3O2(aq) C) H+(aq) + KOH(aq) ÆH2O(l) + K+(aq) D) C2H3O2

- (aq) + H2O (aq) ÆOH- (aq) + HC2H3O2 (aq) E) HC2H3O2 (aq) + OH- (aq) Æ H2O(l) + C2H3O2

- (aq) (HC2H3O2 (aq weakly dissociated acid, major species is HC2H3O2 (aq ) 5. How many grams of NaOH (molar mass= 40.0g) are required to make 0.5L of a pH 13.3 solution?

A) 2.0 g B) 8.0 g C) 3.0 g D) 4.0 g E) 5.0 g

0.7

14 14 13.3 0.70.16pH 7.54 log 7.340.25

[OH ] 10 0.20M0.20 40.0gNaOH 0.5 4.0

pOH pH

moles gL gL mol

� �

� �

�

u

6. Which of the following will have the highest pH?

A) 0.10M HCN Ka= 6.2 x 10-10 B) 0.01M HCN Ka= 6.2 x 10-10( low Ka, lowest concentration of acid, highest pH) C) 0.10M HNO2 Ka= 4.0 x 10-4 D) 0.01M HNO2 Ka= 4.0 x 10-4 E) 0.01M HC2H3O2 Ka= 1.8 x 10-5

7. A student weighs 2.5 x 10–2 mol of a weak monoprotic acid and dissolves it in 350 mL of water. The pH of the resulting solution is 3.05. Calculate the Ka of the acid.

A) 1.3 x 10–2 B) 1.1 x 10–5 C) 7.1 x 10–2 D) 3.3 x 10–5 E) none of these

Chem 126 Common Exam 2 March 21, 2016

4

� � � � � � � �

2

+ -3.05 4

2 3

2.5 10 0.07140.35

pH= 3.05, [H ]=10 8.9 10 I 0.0714 M - ~0

molMolarity ML

HA aq H O aq H O aq A aq

�

�

� �

u

u

� o �

4 4 4

4 4

0C -8.9 10 + 8.9 10 +8.9 10E ~ 0.0714 + 8.9 10 +8.9 10 ignore x

� � �

� �

u u u

u u

> @

4

4 23 5

as (8.9 10 / 0.0714) 100 1.2%,,5%ignore x

(8.9 10 ); 1.1 100.0714a

H O AK

HA

�

� � ��

u u

ª º ª º u¬ ¼ ¬ ¼ u

8. What is the Kb of a base whose 0.294 M solution has a pH of 11.80?

A) 4.9 × 10-7 B) 1.6 × 10-12 C) 2.7× 10-5 D) 1.3 × 10-4 E) 2.1 × 10-2

� � � � � � � �

2.2

2

14 11.8 2.2[ ] 10 0.00630.0063 100 2.1%,5% 0.294

B I 0.294M - ~0

pOHOH M

ignore x

B aq H O aq H aq OH aq

� �

� �

�

u

� o �

0C - x + 0.0063 + 0.0063E ~ 0.294 + 0.0063

> @2

4

+ 0.0063

(0.0063); 1.3 100.294b

BH OHK

B

� ��

ª º ª º¬ ¼ ¬ ¼ u

9. Which of the following reaction(s) are Lewis acid/base reactions?

I. H2O(l) + SO3(g)ÆH2SO4(l) II. AgNO3(aq) + NaCl(aq) ÆAgCl(s) + NaNO3 (aq) III. AlCl3(g) + 6H2O(l) Æ[Al(OH2)6]3+(

aq) + 2Cl-(aq) IV. 2HCl(aq) + Pb(C2H3O2)2 (aq)Æ PbCl2 (s) + HC2H3O2(aq)

A) I and II

Chem 126 Common Exam 2 March 21, 2016

5

B) I, III and IV C) I , II and III D) I, II and IV E) II and III

10. H2PO4⁻ has both a conjugate acid and a conjugate base. What are the conjugate acid/ base of H2PO4⁻ respectively?

A) H3PO4, PO43–

B) H3PO4, HPO42–

C) H2PO4–, HPO4

2– D) HPO4

2–, PO43–

E) HPO42–, H3PO4

Ans B 11. Calculate the [H+] in a solution that is 0.16 M in NaF and 0.25 M in HF. (Ka = 7.2 x 10–4)

A) 7.2 x 10–4 M B) 1.6 M C) 1.1 x 10–3 M D) 0.20 M E) 4.6 x 10–8 M

7.34 8

log ,

0.16pH 7.54 log 7.340.25

[ ] 10 4.6 10

basepH pKaacid

H � �

�

�

� u

12. You wish to prepare an HC2H3O2 buffer with a pH of 4.14. If the pKa of is 4.74, what ratio of C2H3O2⁻/HC2H3O2 must you use?

A) [C2H3O2⁻] is 2 times that of [HC2H3O2] B) [C2H3O2⁻] is 5 times that of [HC2H3O2] C) [C2H3O2⁻] is 4 times that of [HC2H3O2] D) [C2H3O2⁻] is 0.5 times that of [HC2H3O2] E) [C2H3O2⁻] is 0.25 times that of [HC2H3O2]

0.6

log ,4.14 4.74 log

0.6 log , 10 0.25

base basepH pKaacid acid

base baseacid acid

�

� �

�

13. A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH. Assume no volume change upon the addition of base. The Ka for HF is 3.5 × 10-4.

Chem 126 Common Exam 2 March 21, 2016

6

A) 3.63 B) 3.57 C) 3.46 D) 3.34 E) 2.89

� � � � � � � �2

Added NaOH will react with acid HF

0.05 0.375 0.375 0.05 - 0.05

OH aq aq F aq H O lI mol mol molC mol

� �� �

�� 0 +0.050

0 0.325 0.425 mol mol

End mol mol

0.425log 3.46 log 3.570.325a

BpH pKHB

�ª º § · � � ¨ ¸« »© ¹¬ ¼

14. A 1.0 L buffer solution is 0.050 M HC2H3O2 and 0.250 M LiC2H3O2. Which of the following actions will destroy the buffer?

A) adding 0.050 moles of NaOH (will react with 0.050 M HC2H3O2 and neutralize it. No more 0.050 M HC2H3O2, only 0.30 M of C2H3O2

-) B) adding 0.050 moles of HCl C) adding 0.050 moles of LiC2H3O2 D) adding 0.050 moles of HC2H3O2 E) None of the above will destroy the buffer.

15. Consider a solution of 2.0 M HCN and 1.0 M NaCN (Ka for HCN = 6.2 x 10–10). Which of the following

statements is true?

A) The solution is not a buffer because [HCN] is not equal to [CN–]. B) The pH will be below 7.00 because the concentration of the acid is greater than that of the

base. C) [H+] > [OH–] D) The buffer will be more resistant to pH changes from addition of strong acid than of strong

base. E) The buffer will be more resistant to pH changes from addition of strong base than of strong

acid. (greater concentration of acid) 16. A 100.0 mL sample of 0.18 M HClO4 is titrated with 0.27 M LiOH. What is the pH of the solution after the addition of after the addition of 30.0 mL of LiOH?

A) 1.21 B) 2.86

Chem 126 Common Exam 2 March 21, 2016

7

C) 2.00 D) 1.12 E) 0.86

4

4 4

4

0.1 0.18 0.018 (in excess)

0.03 0.27 0.0081

Excess moles 0.018 0.0081 0.0099Final volume after addition 0.13

0.0099 0.0760.13

log[ ]Since

molL moleHClOLmol moleLiOHLHClO molHClO

LmolMolarityofHClO ML

pH HHC

�

u

u

�

�

4is a monoprotic strong acid, pH= -log[0.076]=1.12lO

17. A 100.0 mL sample 0.10 M HCl of is titrated against 50.0 mL of 0.11 M NH3 . What is the resulting pH of the solution? ( Kb of NH3 is 1.8 × 10-5).

A) 3.09 B) 10.91 C) 12.48 D) 1.35 E) 1.52

18. The Ksp for PbF2 is 4.0 x 10–8. If a 0.040 M NaF solution is saturated with PbF2, what is the [Pb2+] in the

solution?

A) 1.0 × 10-6 M B) 6.4 × 10-11 M C) 6.3 × 10-4 M D) 2.5 × 10-5 M E) 1.6 × 10-9 M

� � � � � �22

2+ - 2sp

8 2+

2 excess 0 0.04 -s +s 0.04excess s 0.04K =[Pb ][F ]

4.0 10 [Pb ]

PbF s Pb aq F aq� �

�

�

u 2 2+ 5[0.04] ,[Pb ] 2.5 10 M� u

19. Which of the following compounds will have the highest molar solubility in pure water?

A) Ag2S Ksp = 8 × 10-48

Chem 126 Common Exam 2 March 21, 2016

8

B) CuS Ksp = 1.27 × 10-36 C) PbS Ksp = 9.04 × 10-29 D) Al(OH)3 Ksp = 3 × 10-34 E) ZnS Ksp = 1.6 × 10-24

Al(OH)3Makes 4 ions, Ksp = 27s4, s is (Ksp/27)1/4

Chem 126 Common Exam 2 March 21, 2016

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Part B: 24 points total. 2 questions with multiple parts each worth 12 points Answer all the questions. Show your work.

1. A) (6 PTS) A 0.33-mol sample of a diprotic acid, H2A, is dissolved in 250 mL of water. The Ka1 of this acid is 1.0 x 10–5 and Ka2 is 1.0 x 10–10. Calculate the pH of this solution.

� � � � � � � �2 2 3 I 1.32 - 0 0 C -x + x

H A aq H O aq H O aq HA aq� �� o �

> @2

3 5

2

- -3

+xE (1.32-x) 1.32 (ignoring x) +x +x

x; 1.0 101.32

Solve for [F ], ignoring x, x = 3.63 10 Mlog3

a

H O HAK

H A x

pH

� ��

ª º ª º¬ ¼ ¬ ¼ u�

u

� -3.63 10 2.44u

B) ( 6 PTS) A 10.0 mL solution of 0.1M HCl is added to 25.0 mL of 0.2M HF (Ka = 3.9 x 10-4) solution. Calculate the concentration of F- ions in the solution.

� � � � � � � �2 3

(0.2M 0.025L) New molarity of HF = 0.1430.035

(0.1M 0.010L)New molarity of HCl = 0.02860.035

I 0.143 - 0.0286

ML

ML

HF aq H O aq H O aq F aq� �

u

u

� o �

0C -x + x +xE 0.143-x 0.0286+x +0.0286+x

> @3 4

-

(0.0286+ )[F ]; 3.9 100.143

Solve for [F ], ignoring x, [F ] 0.00195M 0.002MAlternately,do quadratic or successive approximation and solve for [F ] 0.0019 0.002

a

H O FK

HFxx

M M

� � ��

�

�

ª º ª º¬ ¼ ¬ ¼ u�

Chem 126 Common Exam 2 March 21, 2016

10

2.A) (5 PTS) Calculate the molar solubility of Fe(OH)2 in a solution at pH= 10.8. Ksp= 8.16 x10 -17

Fe(OH)2 Fe2+ [OH]-

Initial:

excess +s 0.000631

Change:

- (buffer)

Equilibrium: excess 0.000631

� � � � � �22

14 10.8 3.2[ ] 0.000631M

( ) 2 excess 0 0.000631 -s +s 0.000631excess

pOHOHFe OH s Fe aq OH aq

�

� �

�

�

2+ 2sp

15 2+ 2 2+ 9

s 0.000631K =[Fe ][0.000631]

1.8 10 [Pb ][0.000631] ,[Fe ] 4.5 10 M� �u u

B (5 PTS) A weak acid 20.0 mL of a 0.286M HY is titrated with 0.14M NaOH. (Ka = 6.9 x 10-6) Calculate the pH of the solution after the addition of 20.0 mL of NaOH.

Initial moles of HY= 0.020 L x 0.286M = 0.00572mols Initial moles of NaOH=0.020L x 0.14M = 0.0028 mols

� � � � � � � �-2

First step is the stoichiometric reaction OH A

0.00572 0.0028 ~ 0 HA aq aq aq H O l

I mol mol

�� �

0.0028 - 0.0028 0.00286

0.00292 0 0.00286 Now, you have a buffer soluti

C mol mol molEnd mol mol

��

-on, both [HA] and [A ] are present 0.00286 + log 5.16 log 0.00292

5.15

basepH pKaacid

pH

�

Chem 126 Common Exam 2 March 21, 2016

11

C) (2 PTS) In the titration experiment in 2B, which of following is true at equivalence point?

pH at equivalence point: Acidic Basic Neutral Major species at equivalence point: HY OH- Y-

Chem 126 Common Exam 2 March 21, 2016

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