Test 3 Answer sheet
Spring 2011/ CH116
All answers are in BOLD
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) Of the following equilibria, only __________ will shift to the left in response to a decrease in
volume.
2 SO2 (g) + O2 (g)
A) 2 SO3 (g)
1)
B) 2HI (g)
H2 (g) + I2 (g)
2 HCl (g)
C) H2 (g) + Cl2 (g)
D) N2 (g) + 3 H2 (g)
2 NH3 (g)
E) 4 Fe (s) + 3 O2 (g)
2 Fe2O3 (s)
2) A substance that is capable of acting as both an acid and as a base is __________.
A) saturated
B) conjugated
C) autosomal
D) amphoteric
E) miscible
2)
3) The molar concentration of hydronium ion in pure water at 25°C is __________.
A) 7.00
B) 0.00
C) 1.00
D) 1.0 × 10-7
E) 1.0 × 10-14
3)
4) Of the following acids, __________ is not a strong acid.
A) HCl
B) HNO2
C) HNO3
5) Which one of the following is the weakest acid?
A) HNO2 (Ka = 4.5 × 10-4 )
D) H2 SO4
E) HClO4
4)
5)
B) HF (Ka = 6.8 × 10-4 )
C) HCN (Ka = 4.9 × 10-10)
D) Acetic acid (Ka = 1.8 × 10-5 )
E) HClO (Ka = 3.0 × 10-8 )
6) Which one of the following is a Br nsted-Lowry acid?
A) HNO2
B) CH3 COOH
C) HF
D) (CH3)3 NH+
E) all of the above
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6)
7) In the gas phase reaction below, NH3 is acting as a(n) __________ base but not as a(n) __________
base.
..
N + H+
|
HHH
7)
H+
|
N
|
HHH
A) Lewis, Br nsted-Lowry
B) Arrhenius, Br nsted-Lowry
C) Br nsted-Lowry, Lewis
D) Lewis, Arrhenius
E) Arrhenius, Lewis
8) A solution containing which one of the following pairs of substances will be a buffer solution?
A) KBr, HBr
B) NaI, HI
C) CsF, HF
D) RbCl, HCl
E) none of the above
8)
9) What change will be caused by addition of a small amount of HCl to a solution containing fluoride
ions and hydrogen fluoride?
A) The concentration of fluoride ions will increase as will the concentration of hydronium ions.
B) The fluoride ions will precipitate out of solution as its acid salt.
C) The concentration of hydrogen fluoride will decrease and the concentration of fluoride ions
will increase.
D) The concentration of fluoride ion will decrease and the concentration of hydrogen fluoride
will increase.
E) The concentration of hydronium ions will increase significantly.
9)
10) In a solution, when the concentrations of a weak acid and its conjugate base are equal,
A) the buffering capacity is significantly decreased.
B) the system is not at equilibrium.
C) the -log of the [H+] and the -log of the Ka are equal.
10)
11) Of the following solutions, which has the greatest buffering capacity?
A) 0.821 M HF and 0.217 M NaF
B) They are all buffer solutions and would all have the same capacity.
C) 0.121 M HF and 0.667 M NaF
D) 0.821 M HF and 0.909 M NaF
E) 0.100 M HF and 0.217 M NaF
11)
D) all of the above are true.
12) The conjugate base of HSO4 - is __________.
A) HSO4+
B) SO42C) H3 SO4 +
2
12)
D) H2 SO4
E) OH-
13) What is the pH of an aqueous solution at 25.0 °C that contains 3.98 × 10-9 M hydronium ion?
A) 3.980
B) 5.600
C) 8.400
D) 7.000
E) 9.000
13)
14) The pH of a solution prepared by mixing 45.0 mL of 0.183 M KOH and 65.0 mL of 0.145 M HCl is
__________.
A) 0.744
B) 1.314
C) 7.148
D) 2.923
E) 1.966
14)
ESSAY. Write your answer in the space provided or on a separate sheet of paper.
15) What is the pH of a solution prepared by adding 0.20 mole of formic acid, HCO2H, and 0.25 mole of
sodium formate, NaHCO2, in enough water to make a liter of solution? The K a of formic acid is 1.8
× 10––4.
What is the pH of a solution prepared by adding 0.020 mole of formic acid, HCO2H, and 0.025 mole
of sodium formate, NaHCO2, in enough water to make a liter of solution? The K a of formic acid is
1.8 × 10––4.
What is the pH of a solution prepared by adding 0.60 mole of formic acid and 0.75 mole of
sodium formate in enough water to make a liter of solution?
The K a of formic acid is 1.8 × 10––4.
What is the pH of a solution prepared by adding 0.060 mole of formic acid and 0.075 mole of
sodium formate in enough water to make a liter of solution?
The K a of formic acid is 1.8 × 10––4.
Answer:
This is a solved problem in the book.
The moles of the acid and its conjugate base are given in the buffer solution. As it is one liter
solution the Molarity will be the same number.
Ka is also given.
Just plug all this in the Handerson equation and you can directly get the answer.
16) A 35.0 ml solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH. Calculate the pH after
17.5 mL
34.5 mL
10.0 mL
25.0 mL
of base has been added
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Ka of acetic acid is 1.8 x 10-5
Answer:
This is a homework question.
1. First calculate the moles of Acid CH3COOH and the base NaOH
2. You will find that in each case there is more moles of acid than base. It means all
the base will react and get used up.
The reaction is CH3COOH + Na+ + OH- --> CHCOO- + Na + + H2O
a. For every mole of base that will react an equal amount of CH3COO - will form
b. The excess acid will remain behind and it can be calculated by
Moles of initial acid - moles of NaOH = moles of remaining acid.
3. The mixture after the reaction is over is now a buffer. We can calculate the buffer
with the Handerson - Hasselbach equation.
But this equation requires the concentration , not the amount.
4. Conver the moles to Molarity by usinfg the equation
M = Moles / L
5. Calculate the final amount by adding the volume of the acid and the base in the
teaction. , concert it to liters and use it to calculate the concentration of the remaining
acid CH3COOH and the conjugate base CH3COO-.
6 Ust the Handerson equation to finds out the pH. Ka for the acid is given.
You can use the ICE method to do both the problems but it is a longer .
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