Answer to Question #286483 in Chemistry for khryst

Question

The chloride ion content of a 250.0-mL seawater sample was titrated against 0.1102 M silver nitrate, requiring 13.56 mL to reach end point.

 What is the reaction for the titration analysis?

a.      Ag⁺ + NO₃⁻ + Cl⁻ ⇌ Ag⁺ + NO₃⁻ + Cl⁻

b.     AgCl + NO₃⁻ ⇌ AgNO₃ + Cl⁻

c.      NO₃ + Cl ⇌ NO₃Cl

d.     AgNO₃ + Cl⁻ ⇌ AgCl + NO₃⁻

 

    

  How many moles of chloride ion is present in the sample? (2 points)

a.      1.494 × 10⁻³

b.     2.989 × 10⁻³

c.      7.472 × 10⁻⁴

 

     

 What is the molarity of the chloride ion in the seawater sample? (2 points)

a.      5.976 × 10⁻³

b.     0.01196

c.      2.989 × 10⁻³

 

Answer 1

1.d.     AgNO₃ + Cl⁻ ⇌ AgCl + NO₃⁻

2.n₁ = n₂

С_M1 x V₁ = C_M2 x V₂

n₁ = 0.1102 x 0.01356 = 0.001494 mol

Answer a.      1.494 × 10⁻³

3.C_M = n/V = 0.001494 mol / 0.25 = 0.005976 M

Answer a.      5.976 × 10⁻³