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)₄^2- + 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)₄^2- is found to be 0.012 M, find Kc

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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)₄^2- 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)₄^2-

Initial 2.5 6.5 0

Change -x -4x +x

Equilibr 2.5-x 6.5-4x x

Kc = [Zn(OH)₄^2- ] / [Zn⁺²][OH-]²

[Zn(OH)₄^2- ] = 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

queen_honey_blossom answered 1 year ago

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