Answer to Question #159114 in Chemistry for Dexther Villanueva

Question

A sample is analyzed for chloride by the Volhard method. From the following data, calculate the percentage of chloride present: Weight of sample = 6.0000 g dissolved and diluted to 200 mL Aliquot used = 25.00 mL AgNO3 added = 40.00ml of 0.1234M KSCN for back titration = 13.20ml of 0.0930M

Answer 1

m(sample) = 6.0000 g

V_S = 200 mL

V_A = 25.00 mL

V_excess(AgNO₃) = 40.00 mL

C(AgNO₃) = 0.1234 M

C(KSCN) = 0.0930 M

V(KSCN) = 13.20 mL

Solution:

The Volhard method uses a back titration with potassium thiocyanate to determine the concentration of chloride ions in a solution. Before the titration an excess volume of a silver nitrate solution is added to the solution containing chloride ions, forming a precipitate of silver chloride.

[1]: Ag⁺(aq) + Cl^–(aq) → AgCl(s)

Then the excess silver ions react with the thiocyanate ions to form a silver thiocyanate precipitate. 

[2]: Ag⁺(aq) + SCN^–(aq) → AgSCN(s)

Let's find the volume of the AgNO₃ solution (V_residual) consumed for thiocyanate ions (SCN^–):

C(AgNO₃) × V_residual(AgNO₃) = C(KSCN) × V(KSCN) (according to the equation [2]).

0.1234 M × V_residual(AgNO₃) = 0.0930 M × 13.20 mL

V_residual(AgNO₃) = 0.0930 M × 13.20 mL / 0.1234 M = 9.948 mL = 9.95 mL

V_residual(AgNO₃) = 9.95 mL

Let's find the volume of the AgNO₃ solution (V_reacted) consumed for chloride ions (Cl^–):

V_excess(AgNO₃) = V_reacted(AgNO₃) + V_residual(AgNO₃)

V_reacted(AgNO₃) = V_excess(AgNO₃) - V_residual(AgNO₃) = 40.00 mL - 9.95 mL = 30.05 mL

V_reacted(AgNO₃) = 30.05 mL

Let's find the amount of chloride ions (Cl^–) in an aliquot (V_A = 25.00 mL):

C(Cl^–) × V_A = C(AgNO₃) × V_reacted(AgNO₃) (according to the equation [1]).

n(Cl^–) = C(AgNO₃) × V_reacted(AgNO₃)

n(Cl^–) = 0.1234 M × 0.03005 mL = 0.003708 mol

n(Cl^–) = 0.003708 mol

Let's find the amount of chloride ions (Cl^–) in a solution (V_S = 200 mL):

n_o(Cl^–) = n(Cl^–) × (V_S / V_A)

n_o(Cl^–) = 0.003708 mol × (200 mL / 25.00 mL) = 0.02966 mol

m(Cl^–) = n(Cl^–) × M(Cl^–) = 0.02966 mol × 35.453 g mol^--1 = 1.0515 g

Hence,

w(Cl^–) = [m(Cl^–) / m(sample)] × 100%

w(Cl^–) = [1.0515 g / 6.0000 g] × 100% = 17.525%

w(Cl^–) = 17.525%

Answer: w(Cl^–) = 17.525%.