Question

If you had a 0.650 L solution containing 0.0120 M of Zn2+(aq), and you wished to add enough 1.34 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.

Answer 1

11.6 mL

Step-by-step explanation:

First we need to calculate the number of moles of Zn2+ present in the solution:

`n=V*C\\n_(Zn^(2+))=0.65*0.012=7.8x10^(-3)moles`

As the charge of ion zinc is 2+ and the charge of hydroxide is 1-, we need double moles of NaOH:

`n_(NaOH)=0.0156moles`

As we have the concentration and the moles, we can calculate the volume:

`V=(n)/(C) \\\\V=0.01164L=11.6mL`