126 Version Quiz 3 laude (51635)
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before answering.
001
5.0 points
200 mL of a 2 M solution of sodium chlorite
(NaClO2) is titrated with 500 mL of 0.6 M
HBr. What is the approximate pH of the
resulting solution? HClO2 has a pKa of 2.
1
A 50 mL sample of Ca(OH)2 is titrated to
the equivalence point by 200 mL of an HCl
solution which has a pH of 1. What was the
molarity of the Ca(OH)2 solution before the
addition of any HCl?
1. 0.2 M correct
2. 0.5 M
3. 0.25 M
1. 7
4. 0.1 M
2. 2.5
5. 0.4 M
3. 6
Explanation:
At the equivalence point, nOH = nH+ . Every mole of Ca(OH)2 produces 2 mol of OH .
A pH of 1 corresponds to a molarity of 0.1 for
HCl.
(0.05 L)(2)[Ca(OH)2] = (0.2 L)(0.1 M)
[Ca(OH)2 ] = 0.2 M
4. 2
5. 1.5 correct
Explanation:
Ca
[H+] = Ka
Cb
(0.5 L)(0.6 M)
= 10
(0.2 L)(2 M) (0.5 L)(0.6M)
= 3 102
pH 1.5
004
5.0 points
2
002
5.0 points
How many equations are needed to fully
determine an aqueous system containing the
acid H3 PO4 ?
1. 7
Which of the following assumptions are
made when treating the equation [H+ ] =
(Ka Ca )1/2 as a good approximation of the
proton concentration of a weak acid solution?
I) waters contribution to the proton concentration is negligible
II) the percent ionization is low
III) [H+] = [A]
1. II
2. 4
2. I
3. 5
3. III
4. 6 correct
4. I, II
Explanation:
The system would have an unknown
value for [H+ ], [OH ], [H3 PO4 ], [H2PO4 ],
[HPO4 2 ] and [PO4 3 ].
003
5.0 points
5. II, III
6. I, III
7. I, II, III correct
Version 126 Quiz 3 laude (51635)
Explanation:
In deriving the equation [H+ ] = (Ka Ca )1/2 ,
we ignore the protons that result from autoprotolysis that makes both statements I and
III true, and we also assume that the value
of Ca is large relative to the change in the
concentration of the acid. In other words, we
assume that the percent ioniation is low, so
statement II is true.
005
5. 13
Explanation:
Equal of volumes the titrant and analyte
will be used to reach the equivalance point.
Regardless of the starting volume, at the
equivalance point the volume will be double
the starting value and all of the benzoic acid
will have been converted to benzoate. The
solution will be 0.5 M C6 H5 COO .
5.0 points
Which of the following pairs of solutions
would not result in a buer upon mixing?
1. 1 L of 0.25 M HClO2 ; 1 L of 0.15 M
CsOH
2. 1.5 L of 1 M HCl; 1 L of 2 M CH3 NH2
Kb = Kw /Ka
= 1014 /(5 105 )
= 2 1010
[OH ] = (Kb Cb )1/2
3. 2.5 L of 0.1 M NH4 + ; 1 L of 0.1 M
Ca(OH)2
4. 2 L of 0.1 M LiOH; 1 L of 0.1 M NH3
correct
5. 3 L of 0.2 M NaOH; 4 L of 0.2 M HF
Explanation:
2 L of 0.1 M LiOH mixed with 1 L of 0.1 M
NH3 would simply be a strong base solution,
as the strength of the LiOH would overwhelm
the NH3 and prevent rather than promote
formation of its conjugate acid NH4 + .
006
2
5.0 points
A 1 M solution of NaOH is used to titrate a
1 M solution of C6 H5 COOH (benzoic acid). If
the Ka of C6 H5 COOH is 5 105, what is the
pH of the solution at the equivalence point?
= (2 1010 0.5)1/2
= (1010 )1/2
= 105
pOH = 5
pH = 9
007
5.0 points
One liter of a buer containing 2 M H2 CO3
and 1.5 M NaHCO3 has 24 g of LiOH added
to it. What is the new pH? Assume the Ka1
of H2 CO3 is 2.5 104 .
1. 4 correct
2. 8
3. 11
4. 3
1. 9 correct
5. 6
2. 5
6. 10
3. not enough information
4. 7
Explanation:
1 mol
(24 g LiOH)
24 g
= 1 mol LiOH
Version 126 Quiz 3 laude (51635)
[H+ ] = Ka (Ca /Cb )
= 2.5 104 (1 M/2.5 M)
= 104
pH = 4
008
5.0 points
What would be the pH of a solution prepared from 1 L of H2 O, 120 g of CH3 COOH
and 328 g of NaCH3 COO? Assume the Ka of
acetic acid is 2 105 .
1. 4.4
2. 9
3. 7
4. 9.6
5. 5 correct
Explanation:
(120 g CH3 COOH)
1 mol
60 g
= 2 mol CH3 COOH
2 mol CH3 COOH
= 2 M CH3 COOH
1 L H2 O
(328 g NaCH3COO)
1 mol
82 g
= 4 mol NaCH3COO
4 mol NaCH3COO
= 4 M NaCH3COO
1 L H2 O
[H+ ] = Ka (Ca /Cb )
= 2 105 (2 M/4 M)
= 105
pH = 5
3