A chemical analysis of a water sample (pH 7) yields the following data: Ion Conc

Question

A chemical analysis of a water sample (pH 7) yields the following data:

Ion Concentration (mg/L)

Ca2+ 70

Mg2+ 40

Na+ 10

HCO3 - 150

SO4 2- 120

Cl- 60

a) Calculate the total hardness (in mg/L as CaCO3). Would this water be classified as soft, moderately hard, hard, or very hard?

b) Calculate the carbonate hardness (in mg/L as CaCO3).

c) Calculate the noncarbonate hardness (in mg/L as CaCO3).

d) Calculate the total alkalinity (in mg/L as CaCO3). Would this water be classified as corrosive, stable, or scale-forming?

Explanation / Answer

Dear Student,

table-1 Indication values uses for Hardness and softness

Total Hardness (mg/L as CaCO3)

• Soft: 0-30

• Moderately soft: 30-60

• Moderately hard: 60-120

• Hard: 120-180

• Very hard: >180

a)

We Know, the molar mass of CaCO3,Ca+,Mg+ are 100.1 g/mole, 40.1g/mole, 24.3g/mole respectively.

Therefore molar ratio

MCaCO3/MCa=100.1/40.1=2.5

MCaCO3/MMg=100.1/24.3=4.1

Total water hardness expressed as equivalent of CaCO3 can be calculated with the following formula: Total Permanent Hardness (Mg/L as CaCO3) = 2.5[Ca2+] + 4.1[Mg2+] = 2.5[70] + 4.1[40]=339 Mg/L as CaCO3

Therefore water Cllasified as a very Hard.

b)Calculation Of Bicarbonate Hardness:

Molar mass of HCO-3 is 122.07g/mole

MCaCO3/MHCO3= 100.1/122.07= 0.82

Therefore Bicarbonate Hardness = 0.82[HCO-3]=0.82*150=123 Mg/L as CaCO3.

c) Therefore NonCarbonate Hardness = Total Hardness- Carbonate Hardness

=(339-123)Mg/L as CaCO3

=216 Mg/L as CaCO3

d) Calculating Alkalinity

for Ca+

(70.00 mg Ca{2+} / L) x (1 mmole Ca{2+} / 40.1 g) x (2 meq Ca{2+} / mmole Ca{2+}) x (100 mg CaCO3 / 2 meq) = 174.56 mg/L as CaCO3,

for Mg+

(40.00 mg Mg{2+} / L) x (1 mole Mg{2+} / 24.3 g) x (2 meq Mg{2+} / mmole Mg{2+}) x (100 mg CaCO3 / 2 meq) = 111.088 mg/L as CaCO3=164.6 as CaCO3,

same procedure for rest Ion.

Regards

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