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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.

Respuesta :

Answer:

11.6 mL

Explanation:

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

[tex]n=V*C\\n_{Zn^{2+}}=0.65*0.012=7.8x10^{-3}moles[/tex]

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

[tex]n_{NaOH}=0.0156moles[/tex]

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

[tex]V=\frac{n}{C} \\\\V=0.01164L=11.6mL[/tex]

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