1. Given the following equilibrium constants, calculate the solubility (moles/L) of Ni(OH) 2(s) in a solution that has a fixed [OH - ] of 3.2x10 -7 M Ni(OH) 2(s) K sp = 6x10 -16 NiOH + (aq) β 1 = 1x10 4 Ni(OH) 2(aq) β 2 = 1x10 9 Ni(OH) 3 - (aq) β 3 = 1x10 12 Ni 4 (OH) 4 4+ (aq) β 3 = 2x10 28 The solubility of Ni(OH) 2 is 5.9x10 -3 M. Notice that this answer is dominated by the concentration of Ni 2+ ion, which means that the common ion effect has the biggest influence on the solubility at this concentration of OH - and complex ion formation has no influence on the solubility. How high does the OH - concentration have to be before complex ion formation becomes important?

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1. Given the following equilibrium constants, calculate the solubility (moles/L) of Ni(OH)2(s) in a solution that has a fixed [OH-] of 3.2x10-7M

Ni(OH)2(s) Ksp = 6x10-16

NiOH+(aq) β1 = 1x104

Ni(OH)2(aq) β2 = 1x109

Ni(OH)3-(aq) β3 = 1x1012

Ni4(OH)44+

(aq) β3 = 2x1028

The solubility of Ni(OH)2 is 5.9x10-3 M. Notice that this answer is dominated by the concentration of Ni2+ ion, which means that the common ion effect has the biggest influence on the solubility at this concentration of OH- and complex ion formation has no influence on the solubility. How high does the OH- concentration have to be before complex ion formation becomes important?