Chapter 14: Water Treatment

Problem: 1, Page No:378

In [2]:
# Variables
wt_CaSO4 = 160                  # mg / l

# Solution
hardness = 100 * wt_CaSO4 / 136.
print "The hardness is", "{:.2f}".format(hardness), "mg / L of CaCO3 eqv."

The hardness is 117.65 mg / L of CaCO3 eqv.

Problem: 2, Page No:378

In [4]:
# Variables
wt1 = 9.3                   # mg / L
wt2 = 17.4                  # mg / L
wt3 = 8.7                   # mg / L
wt4 = 12.6                  # mg / L

# Solution
temp_hardness = wt1 * 100 / 146. + wt2 * 100 / 162.
per_hardness = wt3 * 100 / 95. + wt4 * 100 / 136.
total_hardness = temp_hardness + per_hardness

print "Temporary hardness:", "{:.2f}".format(temp_hardness), "mg / L"
print "Total hardness:", "{:.2f}".format(total_hardness), "mg / L"

Temporary hardness: 17.11 mg / L
Total hardness: 35.53 mg / L

Problem: 3, Page No:378

In [5]:
# Variables
wt1 = 32.4                  # mg / L
wt2 = 29.2                  # mg / L
wt3 = 13.5                  # mg / L

# Solution
temp_hardness = wt1 * 100 / 162. + wt2 * 100 / 146.
per_hardness = wt3 * 100 / 136.

print "Temporary hardness:", "{:.2f}".format(temp_hardness), "mg / L"
print "Permanent hardness:", "{:.2f}".format(per_hardness), "mg / L"

Temporary hardness: 40.00 mg / L
Permanent hardness: 9.93 mg / L

Problem: 4, Page No:379

In [3]:
# Variables
i1 = 180            # mg / L, CaCl2
i2 = 210            # mg / L, Ca(NO3)2
i3 = 123            # mg / L, MgSO4
i4 = 90             # mg / L, Mg(HCO3)2

# Solution
i1_req = i1 * 100 / 111.
i2_req = i2 * 100 / 164.
i3_req = i3 * 100 / 120.
i4_req = i4 * 100 / 146.

lime_req = 74 / 100. * (2 * i4_req + i3_req) * 100 / 70. * 10000
soda_req = 106 / 100. * (i1_req + i3_req + i2_req) * 100 / 80. * 10000

print "Lime Required", "{:.1e}".format(lime_req), "mg",
print "=", "{:.1f}".format(lime_req / 10 ** 6), "kg"
print "Soda Required", "{:.1e}".format(soda_req), "mg",
print "=", "{:.1f}".format(soda_req / 10 ** 6), "kg"

Lime Required 2.4e+06 mg = 2.4 kg
Soda Required 5.2e+06 mg = 5.2 kg

Problem: 5, Page No:379

In [8]:
# Variables
wt1 = 32.4                  # mg / L, Ca(HCO3)2
wt2 = 29.29                 # mg / L, Mg(HCO3)2
wt3 = 13.5                  # mg / L, CaSO4

# Solution
wt1_equi = wt1 * 100 / 162.
wt2_equi = wt2 * 100 / 146.
wt3_equi = wt3 * 100 / 136.

temp_hardness = wt1_equi + wt2_equi
perm_hardness = wt3_equi

print "Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is",
print int(temp_hardness), "ppm"
print "Permanent hardness [due to CaSO4] is", "{:.1f}".format(perm_hardness), "ppm"

Temporary hardness [due to Ca(HCO3)2 & Mg(HCO3)2] is 40 ppm
Permanent hardness [due to CaSO4] is 9.9 ppm

Problem: 6, Page No:380

In [9]:
# Variables
v1 = 150                # litres, NaCl

# Solution
v_hardwater = 22500 * v1 / 3 / 0.6 / 58.5

print "The amount of hard water that can be softened using this softner is",
print int(v_hardwater), "litres"

The amount of hard water that can be softened using this softner is 32051 litres

Problem: 7, Page No:380

In [7]:
# Variables
v1 = 30                 # litres, NaCl
w = 1500                # mg / L, NaCl

# Solution
hardness = 45 * 50 / 58.5 * 1000 / 1000
print "Hardness of water is", "{:.2f}".format(hardness), "ppm"

Hardness of water is 38.46 ppm

Problem: 8, Page No:381

In [2]:
# Variables
v1_water = 50               # ml, water
w1_CaCO3 = 1.5              # mg, pure CaCO3
v1_EDTA = 44                # ml, EDTA
v2_EDTA = 40                # ml, EDTA
v2_water = 20               # ml, water

# Solution
EDTA_1ml = v1_water * w1_CaCO3 / v1_EDTA
hardwater_40ml = v2_water * 1.704
total_hardness0 = hardwater_40ml * 1000 / 40
total_hardness1 = total_hardness0 * 0.07

print "Total hardness is", "{:.2f}".format(total_hardness1), "°Cl"

Total hardness is 59.64 °Cl

Problem: 9, Page No:381

In [14]:
# Constants
Fe = 56
S = 32
O = 16
Ca = 40
C = 12

# Solution
hardness100 = Fe + S + O * 4

print "215 ppm of hardness is", "{:.1f}".format(hardness100 * 215 / 100.),
print "ppm of FeSO4"

215 ppm of hardness is 326.8 ppm of FeSO4

Problem: 10, Page No:381

In [1]:
# Variables
v1 = 50.                # ml, hardwater
v2 = 15                 # ml, EDTA
m = 0.01                # M, EDTA

# Solution
M = v2 * m / v1
N = M * 2
S = N * 50 * 1000

print "Molarity of hardness is", M, "M"
print "Normality of hardness is", N, "N"
print "Strength of hardness is", S, "ppm or mg / L"

Molarity of hardness is 0.003 M
Normality of hardness is 0.006 N
Strength of hardness is 300.0 ppm or mg / L

Problem: 11, Page No:382

In [6]:
# Variable
C = 16.5                    # ppm, CO3-2

# Solution
Molarity = C * 10 ** - 6 / 60.

print "Molarity of CO3-2 is", "{:.1e}".format(Molarity), "mol / L"

Molarity of CO3-2 is 2.7e-07 mol / L