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Exam3 Sp 2009 General Chemistry 201 May 12, 2009 Read each question in its entirety and follow the directions described in the question and in this opening statement. If you use a note card as

part of this exam, you will need to staple it to this exam when you turn in the exam to your instructor. Write you name in all papers you turn in. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper number of significant figures. By

signing your signature above you agree that you have worked alone and neither give nor received help from any source. Keep your eyes on your paper at all times!

pX and [X] Relationship

pH = -log [H3O+] pOH = -log [OH-] pKa = -log [Ka]

[H3O+]= 10-pH [OH-]= 10-pOH [Ka]= 10-pKa

Kw: Kw = 1•10-14 @ 25°C Kw = Ka • Kb 14 = pH + pOH Henderson- Hasselbalch Equation; pH = pKa + log [Cb/Ca] or pOH = pKb + log [Ca/Cb]

pKa for various substances: Table 1 will be useful for several problems in this exam Substance Formula pKa1 pKa2 pKa3

i) Phenol, C6H5OH 9.89 ii) Hydrogen cyanide, HCN 9.32 iii) Ammonium, NH4+ 9.26 iv) Hypobromous acid HBrO 8.68 v) Hypochlorous acid HClO 7.52 vi) Hydrogen sulfide H2S 7.02 19.00 vii) Carbonic acid H2CO3 6.37 10.25 viii) Nicotinic acid HC6H4NO2 4.85 ix) Acetic acid HC2H3O2 4.74 x) Barbituric acid HC4C3N2O3 4.01 xi) Lactic Acid HC3H5O3 3.85 xii) Formic acid HCHO2 3.74 xiii) Nitrous acid HNO2 3.34 xiv) Hydrofluoric acid HF 3.17 xv) Chloroacetic acid HC2H2O2Cl 2.85 xvi) Phosphoric acid H3PO4 2.12 7.21 12.38 xvii) Arsenic acid H3AsO3 2.25 7.00 11.52 xviii) Sulfurous acid H2SO3 1.91 7.21 xix) Sulfuric acid H2SO4 Strong 1.99

€

T F =1.8T C + 32

In any aqueous solution Kw is temperature-dependent, the auto-ionization rxn is endothermic, so Kw increases with temperature (°C). °C Kw 10 0.29•10-14 15 0.45•10-14 20 0.68•10-14 25 1.01•10-14 30 1.47•10-14 50 5.48•10-14

Multiple choice (4 pts each). For this exam assume T = 25°C unless otherwise noted. 1 AT 25°C, a 10.0 M solution of LiOH has a pH of: a) 14.00 b) -1.00 c) 15.00 d) No right answer

2 Which solution will have the largest percent ionization, α ? a) 0.01 M HNO2 b) 0.01 M HF c) 0.01 M H2CO3 d) all have the same α

3 Which is strongest acid in water? a) 1.0M HI b) 0.10M HNO3 c) 0.50M HClO4 d) all have the same strength

4 Which salt will not exhibit amphoteric properties? a) LiClO4 b) NaH2PO4 c) NaHSO3 d) all will exhibit amphoterism

5 Which is the strongest electrolyte?

a) HF b) HCH3COO c) HCN d) all are strong electrolyte

6 Which is a diprotic acid?

a) Hydrogen cyanide, HCN b) Phosphoric acid, H3PO4 c) Carbonic acid, H2CO3 d) Ammonia, NH4+

7 Which combination of cation and ion is characteristic of a basic salt? a) Salt formed from anion of strong acid and cation of a strong base b) Salt formed from anion of weak acid and cation of a strong base c) Salt formed from anion of weak acid and the cation NH4

+

d) Salt formed from anion of strong base and cation of a weak acid

8 Which will the following will increase the pH of pure water? a) Increase temperature b) decrease the temperature c) increase the pressure d) pH is constant for water 9 What will happen to someone who hypo-ventilate? a) increase blood pH b) decrease blood pH c) maintain blood pH d) no right answer

10 Which of these gases found in the atmosphere do not lead to acidic rain? a) carbon dioxide b) ozone c) nitrogen dioxide d) none

11 (12pts) Consider the titration curve for chemicals a, b, c, d, e and f. 0.100M HCl is the titrant. Temperature = 25°C.

i) Which among chemicals a, b, c, d, e or f represents the strongest base? Acid-a

ii) Which base has a conjugate acid with a Ka = ~ 1.0 10-4 ? Acid-f

iii) What volume of acid was needed to reach the equivalent point for base b ? According to the graph, the equivalent volume is 50.00 mL iv) If 0.050 M solution of HClO3 is used as the titrant instead of HCl, what

volume of acid will be required to reach the equivalent point for acid a? The volume will be twice as much because the concentration is half as strong. The equivalent volume to reach the equivalent point is therefore 100.0 mL v) True or false, the bases shown in the titration curve (a – f) all have different initial

concentration. Justify your answer. False, since the equivalent point has been reached with 50mL, with the base

concentration the same for all titration, it means the concentration of the acid are all the same.

Titration curves of base with 0.100M HCl. Temperature = 25°C

12 (15 pts) State if you agree or disagree with the following statement, then write a short convincing explanation to support or refute the

statement. i) __ Agree __ Disagree: When the mol of HCN is equal to the mol of NaOH, (equivalent point) during a titration, then the

pH of the solution is 7.0 at 25°C.

ii) __ Agree __ Disagree: The lower the pKa value for an acid, the stronger the acid.

ii) __ Agree __ Disagree: Hydronium (H3O+) concentration in a solution is responsible for pH value of the solution.

iv) __ Agree __ Disagree: The percent ionization, α, of an acid can be adjusted by varying the concentration of the acid.

v) __ Agree __ Disagree: 1.0•10-4M HI is a weaker acid than 1.0 M HF.

13 (12 pts) Place the specie in each of the following groups in order of decreasing strength (strongest to weakest. Justify your answer

i) Base strength: a) HSO3-, b) HCO3-, c) H2PO4- ____ b) HCO3-___ > ___ c) HSO3-___ = ____ a) H2PO4- ___

ii) Acid Strength: a) HBrO2 , b) HFO2, c) HIO2 ____ b) HFO2 ___ > ____ a) HBrO2 ____ > ____ c) HIO2 ___

iii) Acid Strength: a) H2Se b) H2O, c) H2S ____ a) H2Se ____ > ____ c) H2S _____ > ____ b) H2O ___

14 (16 pts) Fill the table below. Use the appropriate Kw for the problem. Use correct number of significant figures.

Substance H3O+ OH- pOH pH pKw Kw

i

Bleach at 25°C 4.4•10-12 2.3•10-3 2.64 11.36 14.00 1.01•10-14

ii

Blood at 98°F 3.98•10-8 7.2198•10-7

3.72•10-7

6.141

6.43

7.40

13.55

13.83 2.87•10-14

15 (16 pts) You want to prepare a buffered solution with a pH = 3.50. (Below are two methods to prepare this solution) i) What mass in gram of NaNO2 must be added to 1.00-L solution of 3.50 M HNO2?

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pH = pka + log Cb

Ca

, Molar Mass(NaNO2 ) = 69.1 gmol

3.50 = 3.34 + log Cb

Ca

= 3.34 + logNO2

-[ ]HNO2[ ]

logNO2

-[ ]3.50[ ]

= 3.50 −3.34 = 0.16

10log

NO2-[ ]

3.50[ ] =100.16 =1.445

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NO2-[ ]

3.50[ ]=1.445, NO2

-[ ] =1.445 (3.50) = 5.059M

Mass NaNO2 = 5.059molL

•1.00L•69.1gmol

Mass NaNO2 =349.6g = 350.g

ii) What mass in gram of NaOH must be added to 1.00-L solution of 3.50 M HNO2 ?

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pH = pka + log Cb

Ca

, Molar Mass(NaOH) = 40.1 gmol

, mol HNO2= 3.50mol

HNO2 + OH− H2O + NO2-

s 3.50mol x - 0R - x - x +xf 3.50 - x 0 x

[c] 3.50 - x1.0 L

0 xL

mol HNO2= 3.50mol

logNO2

-[ ]HNO2[ ]

= log x[ ]3.50 - x[ ]

= 3.50 −3.34 = 0.16

€

10log

x[ ]3.50-x[ ] =100.16 =1.445x[ ]

3.50 - x[ ]=1.445

x =1.44 3.50 - x[ ] = 5.04 −1.44x

2.44x =5.04, x = 2.066

NO2-[ ] = 2.066 = [OH−]

molOH− = molNaOH = 2.066molL

•1.00 L = 2.066mol

Mass NaOH = 2.066mol •40.1 g

mol= 82.83g = 82.8 g NaOH

16 (20pts) Determine if the following solution is acidic, basic or neutral and then calculate the pH of each. Assume T = 25°C. i) 1.00 g of K3PO4 in 1.00 L water.

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MolarMassK3PO4= 212.27 g

mol,

Molarity =1.0g•1 mol

212.27 g•

11.00 L

= 4.71•10−3M

K3PO4 100% → 3K+ + PO43- , [PO4

3-] = 4.71•10−3MpKa3 HPO4

2- = 12.38, pKb1 PO43- = 1.62, Kb1 PO4

3- = 2.40•10-2

PO43- + H2O OH- + HPO4

2-

i 4.71•10−3 1•10-7 0Δ - x +x +x[e] 4.71•10−3 −x x x

Kb1 PO43- = 2.40•10-2 = x2

4.71•10−3 −x

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Simplifying assumption critera :[HA]int

Keq

> 400, ∴Error < 5%

4.71•10−3

2.40•10-2 = 0.196, solve by Quadratic

X2 = 2.40 •10-2(4.71•10−3 −x) → X2 + 2.40•10-2x - 1.131•10-4 = 0

x = -2.40 •10-2 ± (2.40 •10-2)2 −4(1.131•10-4 )2

, x =4.033•10-3M

[OH-] = 4.033•10-3M, pOH = 2.39, pH = 11.61∴ Solution is basic

ii) A mixture containing 2.00 g HF and 1.197g of LiOH in 1000.0 g H2O.

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MolarMassHF = 20.01 gmol

, MolarMassLiOH = 23.95 gmol

,

MolarityHF =2.0g•1 mol

20.01 g•

11.00 L

= 0.100M

MolarityLiOH =1.197g•1 mol

23.95 g•

11.00 L

= 0.0500M

.

iii) A mixture of 1.00 M NH3 and 1.00 M HI in 1.00 L solution.

€

MolarityNH3=1.00M, MolarityHI =1.00M,

The HI will neutralize the NH3 and covert to NH4+

NH4+ + H2O H3O

+ + NH3

i 1.00 1•10-7 0Δ - x +x +x[e] 1.0 −x x x

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Ka NH 4+ = 5.50•10-10 = x2

1.00 −x, x = 2.34 •10-5

[H3O+] = 2.34 •10-5, pH = 4.630

∴ Solution is acidic

iv) A mixture of 0.50 M Ca(OH)2 and 1.00 M HI in 1.00 L solution.

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MolarityCa(OH)2 = 0.50M

MolarityOH- =1.00M, MolarityHI =1.00M, This is a neutral solution

€

[H3O+] = 1.00•10-7, pH = 7.00

∴ Solution is neutral

17 (9 pts) A potassium nitrite salt KNO2 dissociates in water according to the following equation:

KNO2 + H2O (l) → K+ (aq) + NO2- (aq) Kb (NO2

-) = 2.18776 •10-11

What mass of potassium nitrite must be dissolved in 0.5000 liters water, so that the pH is 9.000 ?

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MolarMassKNO2= 85.11 g

molKNO2

100% → K+ + NO2-

pKa HNO2 = 3.34, pKb NO- = 10.66,

pH = 9.00, pOH = 5.00 NO2

- + H2O OH- + HNO2

i W 1•10-7 0Δ -1•10-5 1•10-5 +1•10-5

[e] W−1•10-5 1•10-5 1•10-5

€

Kb NO- = 2.19•10-11 = (1•10-5)2

W −1•10-5

2.19•10-11(W −1•10-5) = 1•10-10

2.19•10-11W −2.19•10-16 = 1•10-10

W = 1•10-10

2.19•10-11 = 4.566M

Mass KNO2 = 4.566molL

•0.500L•85.11g

mol=194.3g

18 (20 pts) Calculate the pH of 40.0mL of 0.500

M nitrous acid HNO2 being titrated with 0.500M LiOH. (Show complete work for full credit)

The pH = 3.85 at 50% equivalence point. 1 Vol LiOH = 20.00ml pH = 3.35 or 3.85 i) What is the pH of the nitrous acid analyte at

0.00% of the equivalent point. 6 Vol LiOH = 00.00ml pH = 1.821 or 2.076 ii) What volume base is necessary to reach the

equivalence point? What is the pH of the solution at this point?

6 Vol LiOH = 40.00ml pH = 8.37 or 8.624 iii) What is the pH after 10.00 mL addition of

LiOH past the equivalence point? 5 Vol LiOH = 50.00ml pH = 12.745 2 Combine the information from above and plot the

titration curve of pH vs. Vol. (mL) LiOH. You should have at least four data point for your curve.

Vol LiOH 0.00 20.00 40.00 50.00 pka =3.35 pH = 1.8 3.35 8.24 12.125 pka =3.85 pH = 2.1 3.85 8.62 12.75

19 (EC 10 pts) Consider the titration of 25.0 mL of 0.200 M HCl with 0.200 M KOH. Calculate the pH after 0.00%, 50.0 %, 100.0 % and

110.0 % of titrant has been added to the analyte. Show work to receive credit.

% Titration 0 % 50% 100% 110% Volume titrant 0.00 12.5 25.0 27.50 pH of solution 0.70 1.18 7.00 12.02

20 (10pts) The Ka of a certain indicator is 2.0 •10-6. The color of HIn is green, and that of In- is red. A few drops of the indicator is added to a HCl solution, which is then titrated against a NaOH solution. At what pH will the indicator change color ?

pH = -log[2.0•10-6] = 5.70