Final answer:
The concentration of Cl− in a 100.0-mL sample of water from a fresh-water aquifer is 226 mg/L.
Step-by-step explanation:
The concentration of Cl− ions in a 100.0-mL sample of water from a fresh-water aquifer can be determined through titration with a known concentration of Hg(NO₃)₂. The titration reaction is:
2Cl−(aq) + Hg(NO₃)₂(aq) → HgCl₂(s) + 2NO₃−(aq)
Given that 6.18 mL of 0.0516 M Hg(NO₃)₂ is required to reach the endpoint, we can first calculate the moles of Hg(NO₃)₂ used:
Molarity (M) = moles/volume (L) → moles = Molarity × Volume
Moles of Hg(NO₃)₂ = 0.0516 M × 0.00618 L = 3.189 × 10⁻´ moles
From the stoichiometry of the reaction, 1 mole of Hg(NO₃)₂ reacts with 2 moles of Cl−. Therefore, the moles of Cl− in the sample are:
Moles of Cl− = 2 × Moles of Hg(NO₃)₂ = 2 × 3.189 × 10⁻´ moles = 6.378 × 10⁻´ moles
The mass of Cl− can be calculated using its molar mass (35.45 g/mol):
Mass of Cl− = 6.378 × 10⁻´ moles × 35.45 g/mol = 0.0226 g
To find the concentration in mg/L, we need to convert the mass to milligrams and the sample volume to liters:
Mass in mg = 0.0226 g × 1000 mg/g = 22.6 mg
Volume in L = 100 mL × 1 L/1000 mL = 0.1 L
Concentration of Cl− in mg/L = mass (mg) / volume (L) = 22.6 mg / 0.1 L = 226 mg/L