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 0.0190 M of Ni3+(aq), and you wished to add enough 1.31 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

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

As per this equation we will need 3 moles of NaOH for every mole of Ni3+

0.25 L 0.0190 M Ni3+ has 0.019 x 0.25 = 0.00475 moles of Ni3+

Now 3 equivalent compared to this is 0.00475 x 3 = 0.01425 moles

If we have a 1.31 M NaOH solution to get 0.01425 moles we will need

1.31 x X = 0.01425

X = 0.01088 L

10.88 mL of 1.31M NaOH

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