Acid Base Chemistry =1.8T C +32

Transcription

Acid Base Chemistry =1.8T C +32
Acid Base Chemistry
Name (last)________________(First)______________________
Practice Exam Acid Base Chemistry
Read all questions before you start. 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 own paper at all times
System
LENGTH
VOLUME
MASS
Temperature
English
1 ft = 12 in
1 gal = 4 qt
1 lb = 16 oz
1 mile = 5280 ft
1 qt = 57.75 in3
1 ton = 2000 lb
T  F = 1.8T  C + 32
1 yd = 3 ft
SI-
1 in = 2.54 cm
1 L = 1.057 qt
1 lb = 453.6 g
English
1 mi = 1.609 km
1 qt = 0.946 L
1 oz = 28.35 g
Misc. info
1 mole = 6.02•1023
pX and [X] Relationship
€
€
pOH = -log [OH-]
[H3O+]= 10-pH
Kw = 1•10-14 @ 25°C
Henderson-Hasselbach Equation
pH = pKa + log [Cb/Ca]
(T  F − 32)
1.8
R = 8.314 J/ mol•K or 0.08206 L•atm/mol•K
pH = -log [H3O+]
Kw
T C =
[OH-]= 10-pOH
pKa = -log [Ka]
[Ka]= 10-pKa
Kw = Ka • Kb 14 = pH + pOH
pOH = pKb + log [Ca / Cb]
Quadratic Equation
x=
i)
−b ± b2 − 4ac
2a
pKa for various substances: Table 1 will be useful for several problems in this exam
Substance
Formula
pKa1
pKa2
pKa3
€
Phenol,
C6H5OH
9.89
In any aqueous solution
Kw is temperature-
ii)
Hydrogen cyanide,
HCN
9.32
iii)
Ammonium,
NH4+
9.26
iv)
v)
vi)
Hypobromous acid
Hypochlorous acid
Hydrogen sulfide
HBrO
HClO
H2S
8.68
7.52
7.02
19.00
increases with
vii)
Carbonic acid
H2CO3
6.37
10.25
temperature.
viii)
Nicotinic acid
HC6H4NO2
4.85
ix)
Acetic acid
HC2H3O2
4.74
x)
Barbituric acid
HC4C3N2O3
4.01
xi)
Lactic Acid
HC3H5O3
xii)
Formic acid
xiii)
dependent, the autoionization rxn is
endothermic, so Kw
°C
10°
0.29•10-14
3.85
15°
0.45•10-14
HCHO2
3.74
20°
0.68•10-14
Nitrous acid
HNO2
3.34
xiv)
xv)
Hydrofluoric acid
Chloroacetic acid
HF
HC2H2O2Cl
3.17
2.85
25°
1.01•10-14
30°
1.47•10-14
xvi)
Phosphoric acid
H3PO4
2.12
7.21
12.38
50°
5.48•10-14
xvii)
Arsenic acid
H3AsO3
2.25
7.00
11.52
1
What is the pH of water at15°C?
a) 13.261
b) 6.631
K w @15°C = 0.45 ⋅10−14
c) 7.000
d) 1.00•10-7
e) none
pK w = −log 0.45 ⋅10−14 = 14.347
[
]
pK w = pH + pOH, pH = 7.17
2€
In a titration experiment 62.50 ml of 0.368 M NaOH is added to a 50.0 ml of 0.460M H2SO4 sample.
What is true about the resulting mixture?
a) The solution is basic
b) The solution is acidic
c) The solution is neutral
d) Not enough information
H2SO4
3€
+
2NaOH

H2O + Na 2SO4
Reaction
mol
[H2SO4 ] = 0.46 ∗50.0ml = 23.0mmol
L
H2SO4 + 2NaOH  H2O + Na+ +
HSO4
-23mmol
+ 23mmol
- 23mmol
+ 23mmol + 23mmol
[NaOH] = 0.368
mol
∗ 62.5ml = 23.0mmol
L
-
HSO - is a weak acid which makes the solution acidic
4
Which acid solution will have the largest percent ionization, α, in water?
a) 1.0 M HClO3
b) 0.01M HClO4
c) 0.001M HCl
d) .001 M HI
e) all the same
All strong acids, all 100% α
4
5
6
7
Which definition of acid-base defines acids as a chemical that increase H+ concentration?
a) Lowry–Bronstead
b) Arrhenius
c) Henderson-Hasselbach
d) Lewis
Which common household product will have the lowest pOH value?
a) Rain water
b) blood
c) baking soda solution
Lowest pOH value will be the strongest base
d) ammonia
e) none
e) Lye
What is the one common structural feature of all bases?
a) it possesses at least a lone pair of electrons
b) it possesses a negative charge
c) it is an anion
d) There is an OH- in its formula
e) none
Which will increase the solubility of BaF2?
a) NH4Cl
b) NaOH
e) none
c) SO3
NH4Cl generates NH4+ which dissociates to produce H3O+, acidic solution.
d) BaCl2
Note: SO3 can also be selected because SO3 will form H2SO4 in water which is acidic.
8
9
10
What conjugate pair is best used to prepare a buffer solution with an effective range of 6.2 – 8.2 ?
a) H3PO4 / NaH2PO4
b) NaH2PO4 / Na2HPO4
c) Na2HPO4 / Na3PO4
d) H2CO3 / NaHCO3
Which acid is also the weakest electrolyte?
a) HI04
b) HBr02
c) HCl02
d) HI02
e) none
d) 2.35 × 10-17 M
e) none
What is the [OH-] for a solution at 25°C that has [H3O+] = 8.23 × 10-2 M
a) 4.26 × 10-5 M
b) 2.35 × 10-11 M
1•10-14 / 8.23 • 10-2 = 1.22 • 10-13 M
c) 4.26 × 10-12 M
e) none
11
Consider the titration curve shown: 0.100M Ca(OH)2 is the titrant. The
analyte s are HA(a), HA(b), HA(c), HA(d) and HA(e).
i) Which titration curve to the right (a - e) represent an acid whose conjugate
possesses the smallest kb?
a) curve a
d) curve d
b) curve b
e) curve e
c) curve c
f) all the same
Smallest Kb is weakest base which is the conjugate of the strongest acid.
ii) What is the acid concentration [H3O+] for HA(c), when 25.0 ml of Ca(OH)2 is
added ?
From curve to right, curve c shows pH = 6 at 25.0 mL, H3O+ = 1•10-6 M
iii) For HA(b), estimate the initial concentration of the solution before any titrant
is added.
Equivalent pt reached at 50.0ml
Titration of 25 mL acid analyte (a – e)
with 0.100 M Ca(OH)2
[OH ] = [H3O ]
−
+
0.100
mol
2mol OH∗50.00ml∗
= 10.0mmol = [OH− ] = [H3O + ]
L
1 mol Ca(OH)2
Concentration of Acid, HA(b)
1
10.0mmol ∗
= 0.400M HA(b)
25.00ml
€ iv) Using the information on the cover page, what is the likely identity of acid-d ?
HA(d), 1/2 eq. vol, pH = pKa = 4.00.
From cover page, acid is Barbituric acid
HC4C3N2O3
4.01
12
3.00 g of lactic acid CH3CH(OH)(COOH) is dissolved in 1.000liter solutions?
i) What are the percent ionization, α , and the pH of this solution?
3.00g HLc = .033M HLc
ΔHLc = 1.4•10 -4 (0.033M) = 2.16 ∑10−3 M
[H3O + ] = ΔHLc = K a (0.033M)
α=
ΔHLc ⎛ 2.16 ∑10−3 M ⎞
= ⎜
⎟ ∗100 = 6.54%
HLc ⎝
.033
⎠
pH = −log 2.16 ∑10−3 M = 2.67
(
)
ii) If 0.400 g of NaOH is added to the solution, what is the pH of the solution? Assume that the volume of 0.400 g
of NaOH is negligible.
€
0.400g of NaOH is 0.01 mol NaOH
This reacts with the 0.0333 M [HLc]
HLc + OH

s 0.0333
0.0100
R - .0100
- 0.0100
-
f
.0233
0
Buffer :
Lc
H2O
0
+ 0.0100
-
0.0100
€
€
Cb
0.010
= 3.85 + log
Ca
0.0233
pH = 3.85 + (−0.367) = 3.48100
pH = pKa + log
13
Place the chemicals for each of the following groups in order of increasing strength weakest (1) to strongest (3).
Justify your answer in as much detail as possible.
i) (Acid strength):
a) HNO2
b) HCN
c) HCH3CO2
see pKa values from front page: HCN (1) < HCH3CO2 (2) < HNO2 (3)
ii) (Acid Strength):
a) HBrO2
b) HIO2
c) HClO2
Electronegativity of middle atom, stronger EN, stronger the acid: HIO2 (1) < HBrO2 (2) < HClO2 (3)
iii) (Base Strength):
a) HAsO42-
b) AsO43-
c) H2AsO4-
H2AsO4- is the strongest acid base ka2 > Ka3. Using conjugate rationale, H2AsO4- is the weakest base:
H2AsO4-(2) < HAsO42-(2) < AsO43-(3) <
14
State if you agree or disagree with the following statement. then write a short convincing explanation to support or
refute it.
i) _X_ Agree __ Disagree:
A pOH of 11.93 for a 0.0085 M monoprotic acid suggest that the acid is a strong acid.
-log (0.0085) = 2.07, pOH = 11.93
ii) __ Agree _X_ Disagree:
When 50.0 mL of 0.100M of NaOH is added to 50.00mL of 0.100M HCl, at 50°C, the solution has a pH = 7.0.
The temperature is not 25°C, so the pH is not 7 when there are equal number of strong acid and strong base.
iii) __ Agree _X_ Disagree
1.0 Molar solution of NH4HCO3 will produce an acidic solution.
NH4HCO3 this salt dissociates to NH4+ and HCO3. This is similar to the extra credit question. The Kb for HCO3 is
the largest
(pKb2 = 7.63) resulting in the solution being basic.
iv) __ Agree _X_ Disagree:
When 1.0 Liter of 1.0 M H2S is mixed with a 1.0 Liter of 1.0 M NaOH , the pH of the solution is the same as the pKa
of H2S.
At the equivalent point the pH is never equal to the pKa. It is half-way to the equivalent point in which pH = pKa.
v) __ Agree _X_ Disagree:
The end point will always occur at the equivalent point during a titration.
If you choose the wrong indicator, the color may change before or after the equivalent point.
15
Calculate the pH of the following solutions:
i) 123.0 mg in 10.0 Liter solution of nicotine (HC6H4NO2)
123.0 mg of nicotine is 1.00 mmol
1.00mmol in 10L is 1.00 ⋅10 -4 M
[H3O + ] = 1.413⋅10−5 ∗1.00 ⋅10 -4
[H3O + ] = 3.76 ⋅10 -5 M
pH = 4.425
€
ii) A mixture containing 10 ml each (total volume is 30 ml) of: H2SO4 (3.00•10-4 mol), HClO (3.00•10-3 mol), and HCN (3.00•10-2 mol)
Among a mixture of strong acids and weak acids,
the [H3O + ] from weak acids will be negligible.
3.00 ⋅10−4 mol
= 1.00 ⋅10−2 M
.030L
pH = 2.00
[H3O + ] =
iii) A mixture of 0.100molHCl solution with 0.025mol LiOH in 100ml solution.
€
HCl
s
+
LiOH 
0.100mol
H2O
+
R - 0.025mol - 0.025mol + 0.025mol
f
LiCl
0.025mol
0.075mol
+ 0.025mol
0
[c] .750M
pH = −log (0.75)
pH = 0.125
€
iv) A mixture of 0.100mol HCl with 0.100 mol NH3in 100ml solution.
This is the equivalence point between the mixture of NH3 . NH3 undergoes hydrolysis to NH4+.
NH4 +
+
H2O
 NH3
+
H3O+
0.100mol
0.100mL
C -x
-
+x
+x
[e]
-
+x
+x
i
1.00 - x
H3O+ =
-
k a ∗1.00 = 5.50 ⋅10−10 ∗1.00
H3O+ = 2.35 ⋅10 -5 M
pH = 4.63
€
16
If an aqueous solution is 50°C, what is [H3O+], [OH-], pH and pOH in which methyl red goes from red to yellow?
Its undissociated form is red and its anionic (dissociated) form is yellow. The pKa for methyl red at 50°C is: pKa = 4.8
Note temperature is 50°C, so Kw = 5.48•10-14
[H3O+ ] ⋅ [Mr-]
,note pK w = 13.26
HMr
Ka
[Mr-]
=
+
[H3O ]
[HMr]
Ka =
[Mr-] = [HMr-] when K a = [H3O+ ]
or pK a = pH = 4.8, pOH = 8.46, pH + pOH = 13.26
[H3O+ ] =1.58⋅10 -5 M,
€
[OH-] = 3.47 ⋅10 -9 M
17
A 40.00 ml solution of 0.100M monoprotic acid is titrated with 0.200 M KOH.
The pH of this monoprotic acid at half-way to the equivalent volume is , pH = 4.78
i) What is the pH of the analyte before the titration begins, (0.00 ml titrant)
ii) What volume of KOH is necessary to reach the equivalent point? What is the pH at the equivalent point?
iii) Calculate the pH after the addition of 50%, 100% and 120% of the amount base to reach the equivalent point.
iv) What is the identity of this monoprotic acid ?
(Sketch the titration curve below on the graph provided) (Use extra paper to show your work)
(See red curve above)
i) pH @ 0.0 Titrant = 2.88
ii) Vol Titrant = 20.00 mL
iii) See graph: 0% pH = 2.88, 50% pH = 4.78, 100% pH = 8.79 pH = 12.10
iv Acetic Acid, pKa = 1.75•10-5
18
Generate a titration curve at 0%, 50%, 100% and 120% when 16.00 ml of 1.00M HBr is titrated with 1.00 M KOH. Show
your complete work and graph the result in the graph above. (See blue curve above)
pH
pOH
0%
0.00
14.0
50%
0.477
13.523
100%
7.00
7.00
120%
12.959
1.041
19
Consider again a new solution again of 9.00 g of
lactic acid CH3CH(OH)(COOH) dissolved in
b) What is the pH change if 100.0 mL of 0.550M
HCl is added to this solution?
1.000 liter solutions?
100mL, 0.550M HCl = 55.0mmol or 5.50⋅10 -2 mol
Lc+ H3 O + 
HLc
+ H2 O
a) If 11.2 gram of NaCH3CH(OH)(COO)
s 0.100mol
0.0550mol
R - 0.055
- 0.0550
f 0.045mol
0
(conjugate base) is added to the solution, what
is the pH of the buffer solution?
g
g
, MWt NaLc = 112.0
mol
mol
9.00g HLc = 0.100mol ⇒ 0.100M
f
VT
MWtHLc = 90.0
[c]
0.045mol
1.100L
0.041
11.2g NaLc = 0.100mol ⇒ 0.100M
pH = pKa + Log
Buffer, [HLc] = [NaLc], pH = pKa = 3.85
Cb
Ca
0.100
+ 0.055
0.155
0
0.155mol
1.100L
0
0.141
= 3.85 + Log
-
0.041
0.141
pH = 3.85 + (-0.5364) = 3.31
ΔpH = 3.31 - 3.85 = - 0.54
€
20
€
State if you agree or disagree with the following statements, then write a convincing explanation to support
or refute the statement.
i) In a strong acid – strong base titration problem, the pH at the equivalent point is always 7.0.
ii) The higher the concentration for an acid the stronger the acid.
ii) A strong base will always have a lower pOH (toward 1) than a weak base.
iv) HA and HB are both weak acids although HA is the stronger of the two. It will therefore take more volume of base
to neutralize 50.0mL of 0.10 M HB than 50.0 mL of 0.10 M HA.
21
i) Calculate the concentration of hydronium ion in pure water at 25°C.
ii) In a titration experiment it was found that a 50.0 ml sample of H2SO4 was completely neutralized by 62.50 ml of
0.368 M NaOH. What is the molarity of the H2SO4 sample ?
iii) What is the hydroxide ion concentration for a solution of pH = 0 ?
iv) How much solvent is required to dilute 100 ml of 16.0 M H2SO4 to 2.0 M?
v) Select the strongest and weakest acid from the list below (SA-WA pair)?
1. SiH4
2. H2S
3. HBr
4. HClO2
vi) Approximately, what is the pH of rain water? Explain.
5. H2O
22
Identify and then write the balance chemical equation for each salt below that will yield a basic solution in water.
i) NH4Br
ii) NaCN
iii) BaF2
iv) NaNO3
Which salt from above will have its solubility increase upon addition of NaOH?
23
The solubility product for copper(II) hydroxide is 4.8•10-20. Calculate the solubility of copper carbonate in
i) aqueous solution and
ii) in 1.00M HCl. Reger13
24
Phenolphthalein is a commonly used indicator that is colorless in the acidic form (pH less than 8.3) and pink in the base
form (pH greater than 10.0). It is a weak acid with a pKa of 8.7. What fraction is in the acid form when the acid color
is apparent? What fraction is in the base form when the base color is apparent?
25
Nitrous acid HNO2, has a Ka of 7.1•10-4. What are [H3O+], [NO2-], [OH-] in 0.500 M HNO2 ?
26
The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of 0.100M NaOH is required to reach the
equivalence point. (See #3 for Ka info.)
i) What is the concentration of the original benzoic acid sample?
ii) What is the pH at the equivalence point?
iii) Calculate the pH after addition of 5.00 ml of NaOH and the pOH after addition of 20.0 ml NaOH.
27
What is the pH of an H2PO4-/HPO42- buffer system in which [HPO42-] is twice [H2PO4-]?
28
What is the pOH and pH of a solution when 10.0 g of sodium benzoate (M.Wt. = 144 g/mol,
Ka = 6.3 • 10-5) is added to 100 ml of 0.10 M Ba(OH)2 and 10 ml of 1M HCl?
29
Lactic acid, CH3CH(OH)(COOH), is formed in muscles as a by-product of muscle's contraction. What is the amount of
lactic acid that is formed (in grams) if the pH level for a 1.00 liter solution reaches 2.66 ?
lactic acid = 90.0 g/mol
pKa = 3.90 , Molar mass of
30
How will the addition of the compound in bold effect (increase, decrease or no change) the solubility of the first compound?
Justify your answer by writing chemical equations and discussing these in terms of LeChatelier Principle.
i) Ca(CH3COO)2 and HCl
ii) MgF2 and HCl
iii) AgCl and NH3
31
Barium carbonate dissolves to some extent in the presence of carbon dioxide.
BaCO3(s) + CO2(g) + H2O(l) → Ba 2+ (aq) + 2 HCO3 - (aq)
i) How is the solubility of barium carbonate affected by the pressure of CO2?
ii) How is the solubility of barium carbonate affected by a decrease in the pH?
Keq = 4.5•10-5
32
A hydrazinium nitros salt N2H5NO2 (MW = 79.0 g/mol) dissociates in water according to the following equation:
N2H5NO2 (s) + H2O (l)  N2H5+ (aq) + NO2 - (aq)
If 39.5 grams of the salt is dissolve in 0.500 liters water, what is the pH of the solution?
Ka nitrous acid (HNO2) = 4.6•10-4
Kb Hydrazine (N2H4) = 1.7•10-6
33
34

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