Problem 6 5 marks] A chemical analysis of a water sample (pH7) yields the follow
ID: 701394 • Letter: P
Question
Problem 6 5 marks] A chemical analysis of a water sample (pH7) yields the following data: lon Concentration (mg/L) 70 25 12 175 134 68 Ca2 Mg Na" HCO SO4 Ch a) Calculate the total hardness (in mg/L as CaCOs). Would this water be classified as soft, moderately hard, hard, or very hard? b) Calculate the carbonate hardness (in mg/L as CaCOs) c) Calculate the noncarbonate hardness (in mg/L as CaCOs). d) Calculate the total alkalinity (in mg/L as CaCO). Would this water be classified as corrosive, stable, or scale-forming?Explanation / Answer
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.