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A 15 mL sample of water from the lemur exhibit required 1.58 mL of a sulfuric acid titrant to reacht he endpoint. Calculate the alkalinity of the exhibit in mg/L of CaCO3 if the titrant conentration was 0.019 M. The molecular weight of calcium carbonate is 100.0869 g/mol.

CaCO3(aq) + H2SO4(aq) --> CaSO4(aq) + H2O(l) + CO2(g)

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Based on the mole ratio of the reaction, the alkalinity of the exhibit is 0.02 mg/L of CaCO3.

What is alkalinity?

Alkalinity is a measure of the hydroxide ion concentration of a solution.

From the equation of the reaction, the mole ratio is 1 :1.

1 mole of the sample reacts with 1 mole of acid.

Moles of titrant = 1.58 mL × 0.019 M = 0.003 mmoles

Moles of samples = 0.003 mmoles

The concentration of sample = 0.003 mmol/15 mL = 0.0002 M

  • Mass concentration = molar concentration × molar

mass concentration = 0.0002 M × 100.0869 g/mol

mass concentration = 0.02 g/mL

Converting to mg/L:

0.02 g/mL × 1000 mg/1000 L = 0.02 mg/L

Therefore, the alkalinity of the exhibit is 0.02 mg/L of CaCO3.

Learn more about alkalinity at: https://brainly.com/question/25312410

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