The equilibrium concentration of Ag⁺ in the solution is : 7.5 * 10⁻⁶ M
Given that :
Ksp for AgIO₃ = 3 * 10⁻⁸
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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