Respuesta :
The concentration of Zn²⁺ = 0.065M
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.
