Create assignment, 48975, Quiz 4, Mar 26 at 2:37 pm
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Msci 18 0013
18:04, general, multiple choice, > 1 min, fixed.
001
What is the H3 O+ concentration of a 0.15 M
solution of NH4 Cl in H2 O at 25◦ C ? (Kb for
NH3 = 1.8 × 10−5 )
1. 9.1 × 10−6 M correct
2. 1.6 × 10−3 M
3. 4.9 × 10−5 M
4. 8.2 × 10−1 M
5. 1.0 × 10−14 M
6. 1.1 × 10−9 M
Explanation:
+
*
NH+
4 ) NH3 + H
Ka =
10−14
Kw
=
Kb
1.8 × 10−5
= 5.56 × 10−10
Since NH+
4 is a weak acid, use the weak acid
equation:
p
[H + ] = (Ka Ca )
q
= (5.56 × 10−10 )(0.15)
= 9.13 × 10−6 M
Msci 18 0718
19:02, general, multiple choice, > 1 min, fixed.
002
The following solutions are mixed in equal
volumes.
1) 0.1 M HBr and 0.1 M KBr
1
2) 0.1 M HBr and 0.1 M KBrO
3) 0.1 M HBrO and 0.1 M KBr
4) 0.1 M HBrO and 0.1 M KBrO
Which will give (a) buffer solution(s)?
1. 2, 3
2. 1, 2
3. 4 correct
4. 3, 4
5. 1
Explanation:
HBr and KBr are a strong acid and its salt,
so this choice can be eliminated.
HBr and KBrO are a strong acid and the
salt of a weak acid.
HBrO and KBr are a weak acid and the salt
of a strong acid.
HBrO and KBrO are a weak acid conjugate
base pair.
Msci 18 0842
19:03, general, multiple choice, > 1 min, fixed.
003
What is the pH of a solution that is 0.010 M
in aqueous NH3 and 0.030 M in NH4 NO3 ?
(Kb [NH3 ] = 1.8 × 10−5 )
1. 5.14
2. 5.22
3. 8.78 correct
4. 8.86
5. 8.92
6. 4.27
7. 9.73
Explanation:
[NH3 ] = 0.010 M
[NH4 NO3 ] = 0.030 M
Create assignment, 48975, Quiz 4, Mar 26 at 2:37 pm
Kb = 1.8 × 10−5
pH = ?
Kw
[NH3 ]
+ log
Kb
[NH+
]
µ
¶4
µ
¶
−14
1 × 10
0.01
= − log
+ log
1.8 × 10−5
0.03
= 8.77815
pH = − log
Msci 19 5019
19:99, general, multiple choice, > 1 min, fixed.
004
Which solution has the highest pH?
1. 0.1 M of KCH3 COO,
Ka (HC2 H3 O2 ) = 1.8 × 10−5
2. 0.1 M of KHCOO,
Ka (HCOOH) = 1.8 × 10−4
3. 0.1 M of KOCl,
Ka (HOCl) = 3.5 × 10−8 correct
4. 0.1 M of KNO2 ,
Ka (HNO2 ) = 4.5 × 10−4
5. 0.1 M of KCl,
Ka (HCl) = very large
2
5. 0.010 M (COOH)2 K1 = 5.9 × 10−2 correct
Explanation:
Msci 18 0500
19:04, general, multiple choice, > 1 min, fixed.
006
The acid form of an indicator is yellow and
its anion is blue. The Ka of this indicator is
1 × 10−5 .
What will be the approximate pH range
over which this indicator changes color?
1. 4 < pH < 6 correct
2. 3 < pH < 5
3. 5 < pH < 7
4. 8 < pH < 10
5. 9 < pH < 11
Explanation:
The pKa of this indicator is 5, so the indicator will change colors around pH 5. Thus
you would expect a color change between pH
4 and pH 6.
Explanation:
Msci 18 0425
18:04, general, multiple choice, > 1 min, fixed.
005
Consider calculations of [H3 O+ ] in each of the
following solutions. Do not go through the
calculations. For which calculations is it not
reasonable to assume that × is much less than
the initial concentration?
The × represents concentration ionized.
(Treat K1 same as K)
Msci 19 0618 alg
19:06, general, multiple choice, > 1 min,
wording-variable.
007
Calculate the pH of the solution resulting
from the addition of 30.0 mL of 0.200 M
HClO4 to 60.0 mL of 0.150 M NaOH.
1. 12.52 correct
2. 12.70
1. O.20 M H2 O2 K = 2.4 × 10−12
3. 7.00
2. 0.010 M HCN K = 4.0 × 10−10
4. 1.30
3. 0.010 M H2 S K1 = 1.0 × 10−7
5. 1.48
4. 1.00 M NH3 K = 1.8 × 10−5
6. 0.82
Create assignment, 48975, Quiz 4, Mar 26 at 2:37 pm
7. 13.18
Explanation:
Here it’s important to find out which of
these two species (HClO4 and NaOH) is in excess. The one that is in excess will determine
the pH of this solution. From the formulas of
the two compounds, you can expect that they
will react in a one-to-one fashion.
So our first order of business will be to determine how many moles of each compound we
have.
For HClO4 , we have
³ 1 L ´ ³ 0.200 mol ´
30.0 mL
1000 mL
1L
= 0.00600 mol HClO4
Likewise, for NaOH, we have
³ 1 L ´ ³ 0.150 mol ´
60.0 mL
1000 mL
1L
= 0.00900 mol NaOH
So when HClO4 and NaOH react, all of the
HClO4 will be consumed (it’s the limiting
reagent) and
0.00900 mol − 0.00600 mol = 0.00300 mol
of NaOH will remain. This 0.00300 mol excess of NaOH will determine the pH of this
solution. The solution now is
30.0 mL + 60.0 mL = 90.0 mL
and, since NaOH is a strong base (i.e., it’s
completely dissociated), it contains 0.00300
mol OH− . [OH− ] is then
0.00300 mol
[OH− ] =
0.0900 L
−
[OH ] = 0.0333 M
which means that the pOH of this solution is
pOH = −log [OH− ]
pOH = −log (0.0333)
pOH = 1.48
However, we wanted pH. We can use the equation that relates pH to pOH to get the pH
pH + pOH = 14
pH + 1.48 = 14
pH = 12.52
Msci 19 0705
3
19:08, general, multiple choice, > 1 min, fixed.
008
At a certain point during a titration, 45.0 mL
of 0.300 M NH3 solution has been added to
30.0 mL of 0.150 M HCl solution.
At this point, the solution is
1. acidic, because not all of the acid has been
neutralized.
2. basic, because excess base has been
added. correct
3. acidic, because of hydrolysis of the resulting salt.
4. basic, because of hydrolysis of the resulting salt.
5. neutral.
Explanation:
0.030 L
× 0.150
1L
= 0.0045 mol
mol acid present =
0.045 L
× 0.300
1L
= 0.0135 mol
mol base present =
Since the number of moles of acid is less
than the number of moles of base, the acid is
fully reacted and excess base has been added.
This makes the solution basic.