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.

11.6 mL

Explanation:

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

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

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

well in order to answer this question you have to provide more information < 3

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).