Question

In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO3)2 solution. 2Cl−(aq) + Hg(NO3)2 (aq) ⟶ 2NO3−(aq) + HgCl2 (s) What is the Cl− concentration in a 0.25-mL sample of normal serum that requires 1.64 mL of 8.35 × 10−4 M Hg(NO3)2(aq) to reach the end point?

Answer 1

Step-by-step explanation:

2Cl−(aq) + Hg(NO₃)₂ (aq) ⟶ 2NO₃−(aq) + HgCl₂ (s)

2 mole 1 mole

1.64 mL of 8.35 x 10⁻⁴ M Hg(NO₃)₂ = 1.64 x 10⁻³ L of 8.35 x 10⁻⁴ M Hg(NO₃)₂

= 1.64 x 10⁻³ x 8.35 x 10⁻⁴ moles of Hg(NO₃)₂

= 13.69 x 10⁻⁷ moles of Hg(NO₃)₂

This will react with 2 x 13.69 x 10⁻⁷ moles of Cl⁻

= 27.38 x 10⁻⁷ moles of Cl⁻

This amount is dissolved in .25 x 10⁻³ L of sample

concentration of Cl⁻

= 27.38 x 10⁻⁷ / .25 x 10⁻³ M

= 109.52 x 10⁻⁴ M