Evaluating an Equilibrium Involving a Complex Ion. PLEASE HELP!!!

Question

You have an aqueous solution of chromium(III) nitrate that you titrate with an aqueous solution of sodium hydroxide. After a certain amount of titrant has been added, you observe a precipitate forming. You add more sodium hydroxide solution and the precipitate dissolves, leaving a solution again. What has happened?

a. The precipitate was sodium hydroxide, which re-dissolved in the larger volume.
b. The precipitate was sodium nitrate, which reacted with more nitrate to produce the soluble complex ion Na(NO3)2−.
c. The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex ion, Cr(OH)4−.
d. The precipitate was chromium hydroxide, which dissolved once more solution was added, forming Cr3+(aq).

Please provide and explanation! Thanks in advance!

Answer 1

This can be explained as-

when sodium hydroxide(aq) reacts with chromium III nitrate (aq) then reaction is as :

Cr(NO3)3 (aq) + 3NaOH (aq) ...........> Cr(OH)3 + 3NaNO3

  In this reaction gelatinous precipitates of green color of Cr(OH)3 are formed .

But when excess of NaOH are added to the Cr(OH)3 then the reaction is as:

  Cr(OH)3 + OH- ...........> Cr(OH)4-(a

the excess of strong base dissolves the green colored precipitates and results in green chromite solution that is Cr(OH)4-

is a soluble complex ion .

from above explanation option "c" would be correct.

hope this helps !