Calculate the solubility at 25 °c of ni(oh), in pure water and in a 0.0130 m naoh solution. you'll find k. data in the aleks data tab. round both of your answers to 2 significant digits. 20 solubility in pure water: solubility in 0.0130 m naoh solution: 602 xs ?

a) in pure water:

⇒ S (25°C) Ni(OH)2 = 2.0 E-5 M

b)  in 0.013 M NaOH:

⇒ S = 9.467 E-11 M

Explanation:

a) Ni(OH)2 ↔ Ni2+  +  2OH-

  S                S               2S

⇒ Ksp = [ Ni2+ ] * [ OH- ]² = 1.6 E-14Ksp value comes from the literature

⇒  Ksp = S * 2S² = 1.6 E-14

⇒ 2S³ = 1.6 E-14

⇒ S = ∛( 1.6 E-14 / 2 )

⇒ S = 2.0 E-5 M

b)  NaOH ↔   Na+  +  OH-

 0.0130 M    0.013     0.013

Ni(OH)2 ↔  Ni2+   +    2OH-

   S               S          2S + 0.0130

⇒ Ksp = [ Ni2+ ] * [ OH- ]²

⇒ Ksp = S * (2S + 0.0130 )² = 1.6 E-14

Due to the small value of the Ksp it can be assumed that "s" is a very small number, so "2s" can be neglected when added to 0.0130.

⇒ 1.6 E-14 = S * ( 0.0130 )²

⇒ S = 9.467 E-11 M

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.

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