Let us assume that Zn(OH)2(s) is completely insoluble, which signifies that the

Question

Let us assume that Zn(OH)2(s) is completely insoluble, which signifies that the precipitation reaction with NaOH(aq) (presented in the transition) would go to completion. If you had a 0.250 L solution containing 0.0140 M of Zn2+(aq), and you wished to add enough 1.32 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

Chemical reaction,

Zn2+ + 2NaOH ---> Zn(OH)2 + 2Na+

So we would required 2 moles of NaOH per mole of Zn2+ to completely precipitate Zn(OH)2

moles of Zn2+ = molarity x volume = 0.0140 M x 0.250 L = 3.5 x 10^-3 mols

So, moles of NaOH needed would be = 2 x 3.5 x 10^-3 = 7 x 10^-3 mols

volume of NaOH solution required = moles/molarity = 7 x 10^-3/1.32 = 5303 ml

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