Problem. The pH 7.5, CaCOვ alkalinity in 332mg/l. Find the value of carbonate and bicarbonate alkalinity and carbonate and bicarbonate concentration?
Solution:
As, pH value <8.2, HCOჳ¯ will exists
So, Carbonate Alkalinity = 0
Now,
Total Alkalinity = Carbonate Alkalinity + Bicarbonate Alkalinity
Given Value, Total Alkalinity =332mg/L
We get, Carbonate Alkalinity = 0
So, Bicarbonate Alkalinity = Total Alkalinity − Carbonate Alkalinity
= 332-0
=332 mg/L
Again,
Ca(HCOჳ)₂ ⇌Ca²+ ➕ 2HCOჳ¯
40+(61*2) 2*61
=162g =122g
[ HCOჳ¯] = ( 122 / 162*1) mole/L = 0.75 mole/L
[ COჳ2¯] = 0 mole/L
Answers :
Carbonate Alkalinity = 0 mg/L
Bicarbonate Alkalinity = 332 mg/l
Carbonate Concentration = 0 mole/L
Bicarbonate Concentration = 0.75 mole/L
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