1 / 1

2 15 11

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 = 6 x10 -16 NiOH + ( aq ) β 1 = 1 x10 4 Ni(OH) 2( aq ) β 2 = 1 x10 9 Ni(OH) 3 - ( aq ) β 3 = 1x10 12

hamish-burton
Download Presentation

2 15 11

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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?

More Related