Question

In an alkalinity test, a sample is titrated with a stand-ardized acid. If 7.5 milliliters of an acid with a knownstrength of 0.02 equivalents per liter (or 0.02 N) areneeded to complete a titration with 60 mL of sample,what is the sample's alkalinity in mg/L as CaCO3?

Answer 1

To calculate the sample's alkalinity in mg/L as CaCO3, use the concept of equivalence point and stoichiometry. Convert the meq of acid to mg/L of CaCO3 using the conversion factor of 50 mg/L per meq.

Step-by-step explanation:

To calculate the sample's alkalinity in mg/L as CaCO3, we need to use the concept of equivalence point and stoichiometry. From the titration, we know that 7.5 mL of 0.02 N acid is required to titrate 60 mL of the sample. This means 0.015 milliequivalents (meq) of acid are needed to neutralize the sample.

Since we want to calculate the alkalinity in mg/L as CaCO3, which is a measure of the carbonate equivalent, we need to convert meq of acid to mg/L of CaCO3. The conversion factor is 50 mg/L per meq of acid. So, the alkalinity of the sample is 0.015 meq/L * 50 mg/L per meq = 0.75 mg/L as CaCO3.

Answer 2

Explanations: