Answer – We are given,
[Zn(NO3)2] = 0.18 M , volume = 10 mL
Volume = 20.0 mL , [NaCl] = 0.26 M
At equilibrium, [ZnCl2] = 0.026 M
We know reaction
  Zn(NO3)2 + 2 NaCl ----->
ZnCl2 + 2 NaNO3
IÂ Â
0.18Â Â Â Â Â Â Â Â Â Â Â Â
0.26Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â 0
CÂ Â
-x              Â
-2x           Â
+x    +2x
E 0.18-x     Â
0.26-2x         Â
0.026Â Â Â Â +2x
So, at equilibrium, x = [ZnCl2] = 0.026 M
So, [Zn(NO3)2] = 0.18-x
                        Â
= 0.18-0.026
                        Â
= 0.154
[NaCl] = 0.26-2x
          Â
= 0.26-2*0.026
     Â
     = 0.208
[NaNO3] = 2x = 2*0.026 = 0.052 M
So,
Kc = [ZnCl2]
[NaNO3]2 / [Zn(NO3)2]
[NaCl]2
    = (0.026)(0.052)2 /
(0.154)(0.208)2
    = 0.0106
