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 0.650 L solution containing 0.0140 M of Ni3+(aq), and you wished to add enough 1.38 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

Here:

M(Ni3+)=0.0140 M

M(OH-)=1.38 M

V(Ni3+)=0.650 L

According to balanced reaction:

3*number of mol of Ni3+ =1*number of mol of OH-

3*M(Ni3+)*V(Ni3+) =1*M(OH-)*V(OH-)

3*0.0140 M *0.650 L = 1*1.38M *V(OH-)

V(OH-) = 0.0198 L

Answer: 0.0198 L

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