A water contains 100.0 mg/L CO32- and 75.0 mg/L HCO3- at a pH of 10. Calculate the alkalinity exactly at 25ºC. Approximate the alkalinity by ignoring [OH-] and [H+].
Question
A water contains 100.0 mg/L CO32- and 75.0 mg/L HCO3- at a pH of 10. Calculate the alkalinity exactly at 25ºC. Approximate the alkalinity by ignoring [OH-] and [H+].
Solution
To calculate the alkalinity, we need to consider the concentrations of CO32-, HCO3-, OH-, and H+. However, since the pH is 10, the concentrations of OH- and H+ are negligible and can be ignored.
First, we need to convert the concentrations from mg/L to moles/L (M). The molar mass of CO32- is approximately 60 g/mol and the molar mass of HCO3- is approximately 61 g/mol.
So, the concentration of CO32- is (100.0 mg/L) / (60 g/mol) = 0.00167 M And the concentration of HCO3- is (75.0 mg/L) / (61 g/mol) = 0.00123 M
Alkalinity is the sum of the concentrations of these species, so:
Alkalinity = [CO32-] + [HCO3-] = 0.00167 M + 0.00123 M = 0.0029 M
To convert this back to mg/L, we multiply by the molar mass of CO32- (since it's the species we're interested in):
Alkalinity = (0.0029 M) * (60 g/mol) = 174 mg/L
So, the alkalinity of the water is approximately 174 mg/L.
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