Question

Equal volumes of 0.140 M AgNO3 and 0.200 M ZnCl2 solution are mixed. Calculate the equilibrium concentrations of Ag+ and Zn2+.

Answer 1

To determine the equilibrium concentrations of Ag+ and Zn2+ after mixing equal volumes of 0.140 M AgNO3 and 0.200 M ZnCl2, use the stoichiometry of the reaction and set up the equilibrium concentrations as x M.

Step-by-step explanation:

To determine the equilibrium concentrations of Ag+ and Zn2+ after mixing equal volumes of 0.140 M AgNO3 and 0.200 M ZnCl2, we need to understand the stoichiometry of the reaction between AgNO3 and ZnCl2. The balanced equation for this reaction is: AgNO3 + ZnCl2 → AgCl + Zn(NO3)2

Since the volumes are additive, the moles of AgNO3 and ZnCl2 are the same. We can use the equation to calculate the moles of AgCl and Zn(NO3)2 formed and then convert them to equilibrium concentrations. If x is the concentration of Ag+ and Zn2+ at equilibrium, then the equilibrium concentrations can be expressed as:[Ag+] = [Zn2+] = x M

where M denotes molar concentration. Since the reaction forms 1 mole of AgCl and 1 mole of Zn(NO3)2 for every mole of AgNO3 and ZnCl2, the equilibrium concentrations can be expressed as:[Ag+] = [Zn2+] = x M

Answer 2

The concentration of Zn²⁺ = 0.065M

Step-by-step explanation:

2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

According to the balanced equation, 2 moles of AgNO₃ reacts with 1 mole of ZnCl₂ to form 2 moles of AgCl and 1 mole of Zn(NO₃)₂.

Molarity of AgNO₃ is 0.14M

Molarity of ZnCl₂ is 0.2M

0.14 M of AgNO₃ would react completely with 0.14/2) = 0.07 M of ZnCl₂, but the ZnCl₂ is more concentrated than that, so ZnCl₂ is in excess and AgNO₃ is the limiting reactant.

Neglecting the Ksp of AgCl:

All of the Ag is in the precipitate so the concentration of Ag⁺ is zero.

(0.2 M ZnCl2 originally) - (0.07 M ZnCl2 reacted) = 0.13 M ZnCl₂ = 0.13 M Zn²⁺ of the original ZnCl₂ solution.

But the original solution has been diluted by the addition of an equal volume of AgNO₃, so the concentration in the final solution is one-half of the original, so:

0.13 M Zn²⁺ / 2

= 0.065 M Zn²⁺

AgCl(s) is also produced as a precipitate so the ions comprising AgCl are removed from solution.