Let us assume that Ni(OH)3(s) is completely insoluble, which signifies that the

Question

Let us assume that Ni(OH)3(s) is completely insoluble, which signifies that the precipitation reaction with NaOH(aq) (presented in the transition) would go to completion. Ni3+(aq)+3NaOH(aq) Ni(OH)3(s)+3Na+(aq) If you had a 0.250 L solution containing 0.0230 M of Ni3+(aq), and you wished to add enough 1.39 M NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH(aq) for the precipitation.

Explanation / Answer

Given equation is

Ni3+(aq) + 3NaOH(aq) ? Ni(OH)3(s) + 3Na+(aq)

Moles of Ni3+ = concentration x volume = 0.0230 M x 0.250 L = 0.00575 mol

ratio of moles of OH- : Ni3+ = 3:1

Hence,

moles of OH- = 3 x moles of Ni3+ = 3 x 0.00575 mol = 0.01725 mol

Given that [ NaOH] = 1.39 M

Hence,

volume of NaOH to be added = number of moles / concentration

= 0.01725 mol / 1.39 M

= 0.0124 L

Therefore,

minimum amount of the NaOH(aq) solution needed to add = 0.0124 L

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