In: Chemistry
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
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