Question

If 845mL of a 2.5x10^-5 mol/L solution of Ni(NO3)2 was mixed with 195mL of a 4.86x10^-2 mol/L solution of KOH, would precipitate form? Calculate the ion product for the potential precipitate. The Ksp of Ni(OH)2(s) is 6.0x10^-16

Answer 1

845ml(2.5x10^-5M Ni(NO3)2 + 195ml(0.0486M KOH)
=>0.845(2.5x10^-5)mole Ni(NO3)2 + 195(0.0486)mole KOH
=>(2.113x10^-5)mole Ni(NO3)2 + (9.48x10^-3)mole KOH
=>(2.113x10^-5)mole Ni^2 + (9.48x10^-3)mole OH^-
=>(2.113x10^-5 mole/1.04 Liter Soln)Ni^2 + (9.48x10^-3 mole OH^-/1.04 Liter Soln)OH^-
=> (2.03x10^-8 M)Ni^+2 + (9.12x10^-6 M)OH^- at mixing ...
=> Qsp = [Ni^+2][OH^-]^2
------ = (2.03x10^-8)(9.12x10^-6)^2
------ = 2x10^-18

Ksp=6x10^-16 > Qsp=2x10^-18 => No ppt forms