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

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.