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
H= 1.0 x 10-6 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 d. 6.0
H= 2.61 x 10-2 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 10 6
x
= 1.34 x 10-3
= -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 c. 3.14
pH = -logH pH = -log H
4 = -log H 3.14 = -log H
H = 10-4 H = - antilog ( 0.14 + 3)
H = 0.72 x 10-3 = 7.2 x 10-4
b. 12.60 d. -1.0
pH = -log H pH = -log H
12.60 = -log H 1.0 = log H
H = - antilog (0.6 + 12) H = antilog 1
H = 0.25 x 10-12 H = 10
= 2.5 x 10-13
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:
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 pH = - log [H+]
pOH = - log [OH-] 6 = - log [H+]
8 = - log [OH-] [H+]= 10-6
[OH-] = -antilog 8
[OH-] = 10-8 M
(conc. H+) x (conc. OH-) = 1.0 x 10-14 1 x 10-6 x (conc. OH-) = 1.0 x 10-14
(conc. OH-) = 1.0 x 10-14 / 1 x 10-6 = 1 x 10-8 M
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 : G0 = -2.303 RT log K
a. If G0= 0, K = 1
If G0 < 0, K is greater Than one
If G0 > 0, K is less Than one
b. Calculate G0 at 2980K for a reaction for which K=
1.60 x 10-4
c. Calculate K at 10000K for a reaction for which G0 =
-12,000 cal
Answer :
b. G0 = -2.303 RT log K
= -2.303 x 1.99 cal/K x 298 K x log (1.60 x 10-4)
= 5189,74 cal
c. G0 = -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
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+]
-1,52 + 0,76 = -0,030 log [( .1 2 )
Zn
conc ]
-0,76 = -0,30 log ( .1 2 )
Zn conc
Log ( .1 2 )
Zn
conc = 25,3
( .1 2 )
Zn
conc = 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 XX0 =
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 X
0 = 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
a. log XX0 =
H
2.10 the rate constant , k, for a reaction can be expressed as a function of temperature by the relation:
ln k = RTEn + 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
2.11 the vapor pressure of water is given as a function of temperature by the relation Log
Answer :
2.12 the fraction, F, of molecul having an energy equal to or greater than the distivation energy , En, is given by the equation:
2.13 the Boltzmann equation for the distribution of molecules among two energy levels is
Answer :
a. E2=0, T=300 k
1 2
n n
= e(e1-e2)/RT = e(0-0)/1.99 x 300 = 1
b. E2=1000 cal, T=300 k
1 2
n n
= e(e1-e2)/RT
= e(0-1000)/1.99 x 300 = e-1.67
= 0.187
c. E2=1000 cal, T=600 k
1 2
n n
= e(e1-e2)/RT
= e(0-1000)/1.99 x 600 = e-o,84
