Question

In: Chemistry

The initial concentrations of Na+ and Cl− in the reaction below are each 0.0843 M. If...

The initial concentrations of Na+ and Cl− in the reaction below are each 0.0843 M. If the initial concentration of NaCl is 0 M and the equilibrium constant is Kc=8.21 under certain conditions, what is the equilibrium concentration of Na+?

NaCl(aq)⇌Na+(aq)+Cl−(aq)

Solutions

Expert Solution

NaCl(aq) Na+(aq) + Cl-(aq)

Equilibrium constant Kc = = 8.21

The solution technique involves ICE box, where 'I' represents Initial, 'C' represents Change, 'E' represents Equilibrium. An empty ICE box looks like:

The complete table is

Here 'x' is unknown. For [Na+] and [Cl-] we represent - x because the amount is going down as reaction proceeds. Thus the amount of [NaCl] goes up till equilibrium reaches. At equilibrium the concentration is shown in the box.

Kc =

  

By solving this quadratic equation,

x= (8.3786+8.3769)/2 and (8.3786-8.3769)/2

= 8.3778M and 0.00085M

The x = 0.00085M ( 0.0843-x is - ve, so neglected)

The equilibrium concentration of Na+ = 0.0834 - 0.00085

= 0.08255M

queen_honey_blossom answered 2 days ago
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