Question

Equal volumes of 0.180 M AgNO₃ and 0.160 M ZnCl₂ solution are mixed. Calculate the equilibrium concentrations of Ag⁺ and Zn²⁺.

Answer 1

Since equal volumes of the two solutions were used, theinitial molar concentrations will be halved.

               [Ag⁺]= 0.18 M / 2 = 0.090 M

               [Cl^-]= 2 (0.16 M) / 2 = 0.16 M

Let's assumbe that the Ag⁺ ions and Cl^-ions react completely to form AgCl (s). Then, we'llre-establish the equilibrium between AgCl, Ag⁺, andCl^-.

                          Ag⁺ (aq)    +     Cl^-(aq)           ->         AgCl(s)

Initiial(M):            0.090               0.16 0

Change(M):        -0.090              -0.090                              +0.090

Final(M):            0                        0.070                              0.090

Now, setting up the equilibrium,

                              AgCl(s)     ->          Ag⁺ (aq)       +           Cl^- (aq)

Initiial(M):               0.090                         0                                0.070

Change(M):              -s                            +s                              +s

Equilibrium (M):       0.090-s                       s                              0.070+ s

Set up the K_sp expression to solve for s.

                 K_sp= [Ag⁺] [Cl^-]

               1.6x 10^-10 = (s) (0.070 + s)

               s= 2.0 x 10^-9 M

               [Ag⁺] = s = 2.0 x 10^-9 M

               [Cl^-] = 0.070 M + s = 0.070 M

               [Zn²⁺]= 0.16 M / 2 = 0.080 M

               [NO^-₃]= 0.18 M / 2 = 0.090 M