Question

A 50.0 mL sample of 0.00200 M AgNO3 is added to 50.0 mL of 0.0100 M NaIO3. What is the equilibrium concentration of Ag in solution

Answer 1

The equilibrium concentration of Ag⁺ in the solution is : 7.5 * 10⁻⁶ M

Given that :

Ksp for AgIO₃ = 3 * 10⁻⁸

Determine the equilibrium concentration of Ag in the solution

First step : Calculate the concentration of Ag⁺ and IO⁻₃ in the solution

[ Ag⁺ ] = ( mmol Ag⁺ / mL solution )

          = ( 50 * 0.00200 / 100 mL ) = 0.001 M

[ IO⁻₃ ] = ( mmol IO⁻₃ / mL solution )

           = ( 50 * 0.0100 / 100 mL )  = 0.005 M

Next determine the Ionic product ( Q )

Q = [ Ag⁺ ] [ IO⁻₃ ]

   = 0.001 * 0.005

   = 5 * 10⁻⁶  

Since the value of Q is > Ksp a precipitate ( IO⁻₃ )  will be formed after the completion of the precipitation reaction

Therefore the concentration of the excess  IO⁻₃  = 0.400 mmol / 100 mL

                                                                                 = 0.004 M

Second step : considering the initial and final concentrations

Initial concentrations ( mol/ L )                  Final concentrations ( mol/L )

[ Ag⁺ ] = 0 M                                                  [ Ag⁺ ]  = x

[ IO⁻₃ ] = 0.004 M                                          [ IO⁻₃ ] = 0.004 + x

Final step : Determine the equilibrium concentration of Ag in the solution

Ksp = 3 * 10⁻⁸ = [ Ag⁺ ]  [ IO⁻₃ ]

                      =  ( x ) ( 0.004 + x )

Therefore x = 7.5 * 10⁻⁶ ( Equilibrium concentration of Ag in the solution )

Hence we can conclude that The equilibrium concentration of Ag⁺ in the solution is : 7.5 * 10⁻⁶ M

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