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 0.150 L solution containing 0.0270 M of Ni3+(aq), and you wished to add enough 1.30 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.

Express the volume to three significant figures and include the appropriate units.

Explanation / Answer

In this case, we can see this as a titration. In the equivalence point, the point where the moles are equals, you can determine the volume to do this.

the reaction is already balanced, so we can see that 1 mole of Ni reacts with 3 moles of NaOH so:
moles NaOH/moles Ni = 3/1
moles OH = 3 moles Ni

And moles = M*V so:
1.30 * V = 0.150 * 0.270 * 3
V = 0.0935 L or simply 93.5 mL

Hope this helps

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