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If 10mL of .18M Zn(NO3)2 is added to 20.0mL of .26M NaCl and concentration of ZnCl2...

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Chemistry

If 10mL of .18M Zn(NO3)2 is added to 20.0mL of .26M NaCl andconcentration of ZnCl2 is found to be .026M. find Kc

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4.1 Ratings (628 Votes)

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

I   0.18             0.26               0           0

C   -x                -2x             +x     +2x

E 0.18-x       0.26-2x           0.026     +2x

So, at equilibrium, x = [ZnCl2] = 0.026 M

So, [Zn(NO3)2] = 0.18-x

                          = 0.18-0.026

                          = 0.154

[NaCl] = 0.26-2x

            = 0.26-2*0.026

            = 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
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