Question

In: Chemistry

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

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

Expert Solution

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

queen_honey_blossom answered 2 years ago

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