12.60 = -log 1.0 = log - Chapter 2
Chapter 2
Problems
In working these problems, the following relationships will be useful:
pH
= -log (conc. H+)
T = temperature 0K = temperature in 0C + 2730
R = gas constanc = 1.99 cal/0K
2.1 calculate the pH of solution which have the following concentration of H+
a. 1.0 x 10-6
c. 3.0 x 10-9
b. 2.61 x 10-2
d. 6.0
Answer :
a. 1.0 x 10-6
c. 3.0 x 10-9
-6
H = 1.0 x 10
H = 3.0 x 10-9
pH = 6
pH = 9 -log 3 = 8.52
= 9 – log 3
= 9 – 0.48
= 8.52
b. 2.61 x 10-2
H = 2.61 x 10-2
d. 6.0
H = 6
pH = - log 6 = -0.78
pH = - log 2.61 10
-2
= 2 – log 2.61
= 2 – 0.42
=1.58
2.2 the concentration of H+ in solution A is 100 times as great as that in solution B.
What is the difference in pH between the two solutions? Which solution has the
greater pH?
Answer :
The concentration of H+ in solution A 100 times is greater than solution B. So, PH
A smaller than PH B because solution A contains more H+ making it more acidic
and PH A smaller.
pH of solution B - pH of solution A = 2.00
2.3 in a 0.10 M solution of acetic acid, the concentration of H+ is given by the expression:
(conc. H+)2 = 1.80 x 10-6 what is pH of this solution?
Answer :
(conc. H+)2 = 1.80 x 10-6
(conc. H+) = 1.80 x10 6
= 1.34 x 10-3
pH =-log H
= -log (1.34 x 10-3)
= 3 -log 1.34
= 3- 0.13
= 2.87
2.4 calculate the concentration of H+ in solutions of the following pH:
a. 4.0
c. 3.14
b. 12.60
d. -1.0
Answer :
a. 4.0
pH = -log H
4 = -log H
H = 10-4
b. 12.60
pH = -log H
12.60 = -log H
H = - antilog (0.6 + 12)
H = 0.25 x 10-12
= 2.5 x 10-13
c. 3.14
pH = -log H
3.14 = -log H
H = - antilog ( 0.14 + 3)
H = 0.72 x 10-3
= 7.2 x 10-4
d. -1.0
pH = -log H
1.0 = log H
H = antilog 1
H = 10
2.5 in a solution saturated with hydrogen sulfide, the following relation holds
(conc. H+)2 x (conc. S2-) = 1 x 10-23
what is the concentration of S2- in a solution of this type which has a pH 0f 4.0?
Answer :
pH = - log H
4 = -log H
H = - antilog 4
H = 10-4
(conc. H+)2 x (conc. S2-) = 1 x 10-23
(10-4)2 x (conc. S2-) = 1 x 10-23
10-8 x (conc. S2-) = 1 x 10-23
(conc. S2-) = 10-23 / 10-8
conc. S2- = 10-15 = 1 x 10-15 M
2.6 in any water solution at 250C :
(conc. H+) x (conc. OH-) = 1.0 x 10-14
the term pOH is defined as:
pOH = - log (conc. OH-)
Making use of either or both of these relations, calculate:
a. The concentration of OH- in a solution of pH 6.0
b. The pOH of a solution in which the concentration of pH is 1 x 10-4
c. The pOH of solution in which the concentration of H+ is 2.0 x 10-3
Answer :
a. pH 6
pOH =14 - 6 =8
pOH = - log [OH-]
8 = - log [OH-]
[OH-] = -antilog 8
[OH-] = 10-8 M
pH = - log [H+]
6 = - log [H+]
[H+]= 10-6
(conc. H+) x (conc. OH-) = 1.0 x 10-14
1 x 10 x (conc. OH-) = 1.0 x 10-14
(conc. OH-) = 1.0 x 10-14 / 1 x 10-6
= 1 x 10-8 M
-6
b. pOH = - log [OH-]
= - log 10-4
=4
c. pH = - log [H+]
= - log 2 x 10-3
= 3 - log 2
pOH = 14 - ( 3 -log 2)
= 11 + log 2
= 11 + 0,3
= 11,3
2.7 the standard free enengy change of a reaction, G ( in calories), is related to the
equilibrium constant, K, by the expression :
G 0 = -2.303 RT log K
a. If G 0= 0, K = 1
If G 0 < 0, K is greater Than one
If G 0 > 0, K is less Than one
b. Calculate G 0 at 2980K for a reaction for which K=
1.60 x 10-4
c. Calculate K at 10000K for a reaction for which G 0 =
-12,000 cal
Answer :
b. G 0 = -2.303 RT log K
= -2.303 x 1.99 cal/K x 298 K x log (1.60 x 10-4)
= -1365,72 (4-log 1,6)
= -1365,72 (3,8)
= 5189,74 cal
c.
-2.303 RT log K
-12000 cal = -2.303 x 1.99 cal/K x 1000 K log K
12000 = 4582.97 log K
Log K = 12000 / 4582.97
= 2.62
K= 416.86 = 4.1686 x 10-2 = 4.2 x 10-2
G 0 =
2.8 the potential, E, for reduction of Zn2+ is given by the equation
E = -0.76 - 0.030 log [(1/(conc. Zn2+)]
a. Calculate E when the concentration of Zn2+ is 1.0 x 10-8
b. Calculate the concentration of Zn2+ when E= -1.52
Answer :
a. E = -0.76 - 0.030 log [(1/(conc. Zn2+)]
= -0.76 - 0.030 log [1/ 1 x 10-8]
= -0.76 - 0.030 log 108
=-0.76 - 0.030 x 8
= -1.00
b. E= -0,76 - 0,030 log [1/conc. Zn2+]
-1,52 = -0,76 - 0,30 log [1/conc. Zn2+]
1
-1,52 + 0,76 = -0,030 log [ (conc.Zn 2) ]
1
-0,76 = -0,30 log (conc.Zn 2)
1
Log (conc.Zn 2) = 25,3
1
(conc.Zn 2)
= antilog 25,3
= 1,99 x 10-25
Conc. Zn2+= 1/ 1,99 x 10-25
= 5 x 10-26 M
2.9 the equation for the rate of radioactive decay is:
log
X0
X
=
H
Where X0 original amount of radioactive material , X is the amount the remaining
after time t, and k is the rate constant:
a. If k = 2.00 x 10-3 / min and X0 = 0.0100 g, what is X when t = 50.0 min
b. How long will it take for the amount of radioactive material to drop
from 1.0 g to 0.10 g if k= 1.0 x 10-2/sec
c. Show that the time required for one half of the radioactive material to
decay must be 0.693/k
Answer :
a. log
Log
X0
=
X
0.002 / milx50.0 min
0.01
=
x
2.303
H
0.01
x
log
0.01
x
= o.o4342
= antilog 0.43
= 1.104
X= 9.05 x 10-3 g
b. log
ln
X0
X
1g
0.1g
=
=
H
0.01 / sec .t
2.303
ln 10 = 1.0 x 10-2 / sec t
2.303= 1.0 x 10-2 / sec t
t= 2.303 x 10-2 / sec
c. Log
2
1
t 1/2 =
=
kt1 / 2
2.303
(2.303)(0.3010)
k
t 1/2 = 0.693/k
2.10 the rate constant , k, for a reaction can be expressed as a function of temperature by
the relation:
ln k =
En
RT
+B
Where En is the energy of activation and B is a constant. Calculate k at 250C for a
reaction for which B = 2.10 and En = 1.0 x 10-4 cal
Answer :
ln k =
ln k =
ln k =
En
RT
+B
1.0 x10 4
+
1.99cal / k .298k
1.0 x10 4
+ 2.10
593.02
2.10
ln k = -14.76
k = 3.88 x 10-7
2.11 the vapor pressure of water is given as a function of temperature by the relation
Log
p2
p1
=
10.200(T2 T1 )
2.303 RT2T1
Where P2 and P1 are the vapor preassure at temperatures T2 and T1 respectively. The
vapor pressure of water at 1000C is 760mmHg.
Answer :
a. What is the vapor pressure of water at 250C ( 298 k)
p2
p1
Log
Log
Log
Log
p2
p1
=
10.200(T2 T1 )
2.303RT2T1
10.2
( 1/T1 - 1/ T2)
2.303
760
10.2
P1 = 2.303 x1.99 (1/373
10.2
760 - log P1 = 4.58
=
- 1/298)
b. At what temperature will the vapor pressure of water be 1.2 x 10-3
log
760
1200
=
log 0.63 =
10.2
2.303 x1.99
10.2
(298-T1/
4.58
(1/T1- 1/298)
298T1)
-0.2 = 2.227 (298-T1/ 298T1)
-0.2 =
663.646 2.227T1
298T1
-59.6 T1 = 663.646 - 2.227 T1
-59.6 T1 + 2.227 T1 = 663.646
-57.373 T1 = 663.646
T1 = -11.56722
2.12 the fraction, F, of molecul having an energy equal to or greater than the
distivation energy , En, is given by the equation:
f = e-En/RT
What is f when En = 1.00 x 10-4cal, T= 298k
Answer :
f = e-En/RT
= e-1.00 x 10-4 cal/ 1.99 cal/k. 298k
= e-16.86
ln = -16.86
2.303 log x = -16.86
Log x = -7.32
x = antilog -7.32
x = 4.77 x 10-8
2.13 the Boltzmann equation for the distribution of molecules among two energy
levels is
n2
n1
= e(e1-e2)/RT
Where n1 and n2 are the number of molecules in the levels whose energies are
E2 and E1 respectively. Calculate the ratio n2/n1 whwn E1 =0 and
Answer :
a. E2=0, T=300 k
n2
n1
= e(e1-e2)/RT
= e(0-0)/1.99 x 300
=1
b. E2=1000 cal, T=300 k
n2
n1
= e(e1-e2)/RT
= e(0-1000)/1.99 x 300
= e-1.67
= 0.187
c. E2=1000 cal, T=600 k
n2
n1
= e(e1-e2)/RT
= e(0-1000)/1.99 x 600
= e-o,84
= 0.433
Problems
In working these problems, the following relationships will be useful:
pH
= -log (conc. H+)
T = temperature 0K = temperature in 0C + 2730
R = gas constanc = 1.99 cal/0K
2.1 calculate the pH of solution which have the following concentration of H+
a. 1.0 x 10-6
c. 3.0 x 10-9
b. 2.61 x 10-2
d. 6.0
Answer :
a. 1.0 x 10-6
c. 3.0 x 10-9
-6
H = 1.0 x 10
H = 3.0 x 10-9
pH = 6
pH = 9 -log 3 = 8.52
= 9 – log 3
= 9 – 0.48
= 8.52
b. 2.61 x 10-2
H = 2.61 x 10-2
d. 6.0
H = 6
pH = - log 6 = -0.78
pH = - log 2.61 10
-2
= 2 – log 2.61
= 2 – 0.42
=1.58
2.2 the concentration of H+ in solution A is 100 times as great as that in solution B.
What is the difference in pH between the two solutions? Which solution has the
greater pH?
Answer :
The concentration of H+ in solution A 100 times is greater than solution B. So, PH
A smaller than PH B because solution A contains more H+ making it more acidic
and PH A smaller.
pH of solution B - pH of solution A = 2.00
2.3 in a 0.10 M solution of acetic acid, the concentration of H+ is given by the expression:
(conc. H+)2 = 1.80 x 10-6 what is pH of this solution?
Answer :
(conc. H+)2 = 1.80 x 10-6
(conc. H+) = 1.80 x10 6
= 1.34 x 10-3
pH =-log H
= -log (1.34 x 10-3)
= 3 -log 1.34
= 3- 0.13
= 2.87
2.4 calculate the concentration of H+ in solutions of the following pH:
a. 4.0
c. 3.14
b. 12.60
d. -1.0
Answer :
a. 4.0
pH = -log H
4 = -log H
H = 10-4
b. 12.60
pH = -log H
12.60 = -log H
H = - antilog (0.6 + 12)
H = 0.25 x 10-12
= 2.5 x 10-13
c. 3.14
pH = -log H
3.14 = -log H
H = - antilog ( 0.14 + 3)
H = 0.72 x 10-3
= 7.2 x 10-4
d. -1.0
pH = -log H
1.0 = log H
H = antilog 1
H = 10
2.5 in a solution saturated with hydrogen sulfide, the following relation holds
(conc. H+)2 x (conc. S2-) = 1 x 10-23
what is the concentration of S2- in a solution of this type which has a pH 0f 4.0?
Answer :
pH = - log H
4 = -log H
H = - antilog 4
H = 10-4
(conc. H+)2 x (conc. S2-) = 1 x 10-23
(10-4)2 x (conc. S2-) = 1 x 10-23
10-8 x (conc. S2-) = 1 x 10-23
(conc. S2-) = 10-23 / 10-8
conc. S2- = 10-15 = 1 x 10-15 M
2.6 in any water solution at 250C :
(conc. H+) x (conc. OH-) = 1.0 x 10-14
the term pOH is defined as:
pOH = - log (conc. OH-)
Making use of either or both of these relations, calculate:
a. The concentration of OH- in a solution of pH 6.0
b. The pOH of a solution in which the concentration of pH is 1 x 10-4
c. The pOH of solution in which the concentration of H+ is 2.0 x 10-3
Answer :
a. pH 6
pOH =14 - 6 =8
pOH = - log [OH-]
8 = - log [OH-]
[OH-] = -antilog 8
[OH-] = 10-8 M
pH = - log [H+]
6 = - log [H+]
[H+]= 10-6
(conc. H+) x (conc. OH-) = 1.0 x 10-14
1 x 10 x (conc. OH-) = 1.0 x 10-14
(conc. OH-) = 1.0 x 10-14 / 1 x 10-6
= 1 x 10-8 M
-6
b. pOH = - log [OH-]
= - log 10-4
=4
c. pH = - log [H+]
= - log 2 x 10-3
= 3 - log 2
pOH = 14 - ( 3 -log 2)
= 11 + log 2
= 11 + 0,3
= 11,3
2.7 the standard free enengy change of a reaction, G ( in calories), is related to the
equilibrium constant, K, by the expression :
G 0 = -2.303 RT log K
a. If G 0= 0, K = 1
If G 0 < 0, K is greater Than one
If G 0 > 0, K is less Than one
b. Calculate G 0 at 2980K for a reaction for which K=
1.60 x 10-4
c. Calculate K at 10000K for a reaction for which G 0 =
-12,000 cal
Answer :
b. G 0 = -2.303 RT log K
= -2.303 x 1.99 cal/K x 298 K x log (1.60 x 10-4)
= -1365,72 (4-log 1,6)
= -1365,72 (3,8)
= 5189,74 cal
c.
-2.303 RT log K
-12000 cal = -2.303 x 1.99 cal/K x 1000 K log K
12000 = 4582.97 log K
Log K = 12000 / 4582.97
= 2.62
K= 416.86 = 4.1686 x 10-2 = 4.2 x 10-2
G 0 =
2.8 the potential, E, for reduction of Zn2+ is given by the equation
E = -0.76 - 0.030 log [(1/(conc. Zn2+)]
a. Calculate E when the concentration of Zn2+ is 1.0 x 10-8
b. Calculate the concentration of Zn2+ when E= -1.52
Answer :
a. E = -0.76 - 0.030 log [(1/(conc. Zn2+)]
= -0.76 - 0.030 log [1/ 1 x 10-8]
= -0.76 - 0.030 log 108
=-0.76 - 0.030 x 8
= -1.00
b. E= -0,76 - 0,030 log [1/conc. Zn2+]
-1,52 = -0,76 - 0,30 log [1/conc. Zn2+]
1
-1,52 + 0,76 = -0,030 log [ (conc.Zn 2) ]
1
-0,76 = -0,30 log (conc.Zn 2)
1
Log (conc.Zn 2) = 25,3
1
(conc.Zn 2)
= antilog 25,3
= 1,99 x 10-25
Conc. Zn2+= 1/ 1,99 x 10-25
= 5 x 10-26 M
2.9 the equation for the rate of radioactive decay is:
log
X0
X
=
H
Where X0 original amount of radioactive material , X is the amount the remaining
after time t, and k is the rate constant:
a. If k = 2.00 x 10-3 / min and X0 = 0.0100 g, what is X when t = 50.0 min
b. How long will it take for the amount of radioactive material to drop
from 1.0 g to 0.10 g if k= 1.0 x 10-2/sec
c. Show that the time required for one half of the radioactive material to
decay must be 0.693/k
Answer :
a. log
Log
X0
=
X
0.002 / milx50.0 min
0.01
=
x
2.303
H
0.01
x
log
0.01
x
= o.o4342
= antilog 0.43
= 1.104
X= 9.05 x 10-3 g
b. log
ln
X0
X
1g
0.1g
=
=
H
0.01 / sec .t
2.303
ln 10 = 1.0 x 10-2 / sec t
2.303= 1.0 x 10-2 / sec t
t= 2.303 x 10-2 / sec
c. Log
2
1
t 1/2 =
=
kt1 / 2
2.303
(2.303)(0.3010)
k
t 1/2 = 0.693/k
2.10 the rate constant , k, for a reaction can be expressed as a function of temperature by
the relation:
ln k =
En
RT
+B
Where En is the energy of activation and B is a constant. Calculate k at 250C for a
reaction for which B = 2.10 and En = 1.0 x 10-4 cal
Answer :
ln k =
ln k =
ln k =
En
RT
+B
1.0 x10 4
+
1.99cal / k .298k
1.0 x10 4
+ 2.10
593.02
2.10
ln k = -14.76
k = 3.88 x 10-7
2.11 the vapor pressure of water is given as a function of temperature by the relation
Log
p2
p1
=
10.200(T2 T1 )
2.303 RT2T1
Where P2 and P1 are the vapor preassure at temperatures T2 and T1 respectively. The
vapor pressure of water at 1000C is 760mmHg.
Answer :
a. What is the vapor pressure of water at 250C ( 298 k)
p2
p1
Log
Log
Log
Log
p2
p1
=
10.200(T2 T1 )
2.303RT2T1
10.2
( 1/T1 - 1/ T2)
2.303
760
10.2
P1 = 2.303 x1.99 (1/373
10.2
760 - log P1 = 4.58
=
- 1/298)
b. At what temperature will the vapor pressure of water be 1.2 x 10-3
log
760
1200
=
log 0.63 =
10.2
2.303 x1.99
10.2
(298-T1/
4.58
(1/T1- 1/298)
298T1)
-0.2 = 2.227 (298-T1/ 298T1)
-0.2 =
663.646 2.227T1
298T1
-59.6 T1 = 663.646 - 2.227 T1
-59.6 T1 + 2.227 T1 = 663.646
-57.373 T1 = 663.646
T1 = -11.56722
2.12 the fraction, F, of molecul having an energy equal to or greater than the
distivation energy , En, is given by the equation:
f = e-En/RT
What is f when En = 1.00 x 10-4cal, T= 298k
Answer :
f = e-En/RT
= e-1.00 x 10-4 cal/ 1.99 cal/k. 298k
= e-16.86
ln = -16.86
2.303 log x = -16.86
Log x = -7.32
x = antilog -7.32
x = 4.77 x 10-8
2.13 the Boltzmann equation for the distribution of molecules among two energy
levels is
n2
n1
= e(e1-e2)/RT
Where n1 and n2 are the number of molecules in the levels whose energies are
E2 and E1 respectively. Calculate the ratio n2/n1 whwn E1 =0 and
Answer :
a. E2=0, T=300 k
n2
n1
= e(e1-e2)/RT
= e(0-0)/1.99 x 300
=1
b. E2=1000 cal, T=300 k
n2
n1
= e(e1-e2)/RT
= e(0-1000)/1.99 x 300
= e-1.67
= 0.187
c. E2=1000 cal, T=600 k
n2
n1
= e(e1-e2)/RT
= e(0-1000)/1.99 x 600
= e-o,84
= 0.433