In a common medical laboratory determination of the concentration of chloride ion in blood serum, a serum sample is titrated with a hg(no3)2 solution
2cl- + hg(no3)2 ? 2no3- + hgcl2
what is the cl concentration in a 0.25-ml sample of normal serum that requires 1.46 ml of 5.25 x 10-4 m hg(no3)2 to reach the endpoint?

[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

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