Chapter 5 FE EXAM FORMATTED PROBLEMS
5-1.
The pH of a finished water from an excess lime softening process is 11.24. What volume
of 0.0200 N sulfuric acid, in millimeters, is required to neutralize 1.00 L of the finished
water? Assume the buffering capacity is zero.
Assume only OH is present, then from Eq. 5-29
pOH  14.00  11.24  2.76
OH   10 2.76  1.74 10 3 moles / L
Recognizing that n=1 for OH and that normality = molarity and that
N x ml = N x ml
Then,
(1.74 10 3 )(1,000 ml )  (0.0200 N )(ml of acid)
ml of acid  86.89 or 87 ml
5-2.
What pH is required to precipitate all but 0.20 mg/L of the iron from raw water with an
Fe3+ concentration of 2.1 mg/L? Assume the temperature is 25°C. The reaction is
Fe3+ + 3OH- ↔ Fe(OH)3 ↓ and pKs= 38.57
GMW of Fe = 55.847
Ks= 10-38.57 = 2.69 x 10-39
Using Eq. 5-21
Ks = [Fe][OH]3
1/ 3
39


OH    2.69 10 6 
 3.58  10 
 9.09  10 12 moles / L
pOH   log( 9.09  10 12 )  11.04
pH  14.00  11.04  2.96
5-3.
Estimate the pH of water that contains 0.6200mg/L of carbonic acid. Assume that
[H+] = [HCO3-] at equilibrium and that the dissociation of water may be neglected. The
water temperature is 25°C, pKa1 = 6.35 and pKa2 = 10.33. GMW of H2CO3 = 62 g/mol
Converting mg/L to moles/L
0.6200 mg / L
 1.00  10 5 mole / L
62 g / mol 1000 mg / g 
Using Eq. 5-33 to solve the equilibrium expression for Ka1
K a1 
[ H  ][ HCO3 ]
 10 6.35  4.467  10 7
[ H 2 CO3 ]
Assuming that [H+]=[HCO3-] then,
[ H  ]2
1.00  10 5
[ H  ]2  4.467  10 12
4.467  10 7 
[ H  ]  2.113  10 6
pH   log( 2.113  10 6 )  5.67
5-4.
Estimate the approximate alkalinity, in mg/L as CaCO3, of water with a carbonate ion
concentration of 17.0 mg/L and a bicarbonate ion concentration of 111.0 mg/L.
Calculate the equivalent weight
CaCO3 
100
 50equiv
2
60
 30equiv
2
61
HCO3 
 61equiv
1
Convert to mg/L as CaCO3 using Eq. 5 - 40
CO3 
 50 
CO3  (17 mg / L)   28.33mg / L as CaCO3
 30 
 50 
HCO3  (111mg / L)   90.98mg / L as CaCO3
 6 
Calculate approximat e alkalinity remembering that the 2 multiple for CO 32- has already been accounted for
Alk  28.33  90.98  119.32 or 119 mg/L as CaCO3