Answer:
[Cl⁻] = 6.13 x 10⁻³ M
Explanation:
Hg(NO₃)₂ + 2Cl⁻ ⇒ 2NO₃⁻ + HgCl₂
The amount of Hg(NO₃)₂ that was used to titrate is calculated:
(1.46 mL)(5.25 x 10⁻⁴ mol/L) = 7.665 x 10⁻⁴ mmol
The molar ratio between Hg(NO₃)₂ and Cl⁻ is used to find mmol of Cl⁻ that must have been present in the same:
(7.665 x 10⁻⁴ mmol Hg(NO₃)₂) x (2Cl⁻ / Hg(NO₃)₂) = 1.533 x 10⁻³ mmol Cl⁻
To find the concentration of Cl⁻, the amount in moles is divided by the volume:
[Cl⁻] = (1.533 x 10⁻³ mmol Cl⁻) / (0.25 mL) = 6.13 x 10⁻³ M
