Question

Equal volumes of 0.180 M AgNO3 and 0.140 M ZnCl2 solution are mixed. Calculate the equilibrium concentrations of Ag+ and Zn2+.

Answer 1

The balance chemical equation would for this reaction would be:

2 AgNO3 + ZnCl2 → 2 AgCl + Zn(NO3)2

As can be seen from the stoichiometry, 0.180 M AgNO3 would react completely with 0.180x (1/2) = 0.90 M ZnCl2, but the ZnCl2 is more concentrated than that, so ZnCl2 is in excess and AgNO3 is the limiting reactant.

Ksp of AgCl is very small which can be neglected, so
all of the Ag is in the precipitate so the concentration of Ag+ is zero.

Now for ZnCl2:
(0.140 M ZnCl2 originally) - (0.090 M ZnCl2 reacted)

= 0.050 M ZnCl2

= ie 0.050 M Zn{2+} of the original ZnCl2 solution.
But the original solutionof ZnCl2 has been diluted by the addition of an equal volume of AgNO3, so the concentration of the final solution is one-half of the original, so:

[Zn2+] = 0.050 M / 2

= 0.025 M