Question

In: Chemistry

The reaction Zn(NO3)2 + 4NaOH ---> Zn(OH)42- + 2NO-3 + 4Na+ If 25 mL of a...

The reaction

Zn(NO3)2 + 4NaOH ---> Zn(OH)42- + 2NO-3 + 4Na+

If 25 mL of a 0.10 M Zn(NO3)2 is added to 10.0 mL of 0.65 M NaOH and at equilibrium the concentration of  Zn(OH)42- is found to be 0.012 M, find Kc

Solutions

Expert Solution

If 25 mL of a 0.10 M Zn(NO3)2 is added to 10.0 mL of 0.65 M NaOH and at equilibrium the concentration of  Zn(OH)42- is found to be 0.012 M, find Kc

Initial moles of Zn+2 = Molarity X volume = 0.1 x 25 mL = 2.5 mmoles

Initial moles of OH- = Molarity X volume = 0.65 X 10 = 6.5 mmoles

Total volume = 25 + 10 = 35mL

               Zn+2+ 4OH- ---> Zn(OH)42-

Initial        2.5        6.5         0

Change      -x        -4x       +x

Equilibr      2.5-x     6.5-4x    x

Kc = [Zn(OH)42- ] / [Zn+2][OH-]2

[Zn(OH)42- ] = mmoles / Volume = x / 35 = 0.012 M , x = 0.42

[Zn+2] = mmoles / Volume = 2.5-x / 35 = 0.059

[OH-] = mmoles / Volume = 6.5-4x / 35 = 0.138

Kc = 0.059 X (0.138)4 / 0.012 = 0.00178 = 1.78 X 10-3


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