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?1) The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex ion, Cr(OH)4?.2) The precipitate was chromium hydroxide, which dissolved once more solution was added, forming Cr3+(aq).3) The precipitate was sodium nitrate, which reacted with more nitrate to produce the soluble complex ion Na(NO3)2?.4) The precipitate was sodium hydroxide, which re-dissolved in the larger volume.

Answer 1

Answer:

The precipitate was chromium hydroxide, which then reacted with more hydroxide to produce a soluble complex, Cr(OH)4

Explanation:

The following reaction takes place when chromium(III) nitrate reacts with NaOH:

$Cr(NO)_{3}$ +3 NaOH → $Cr(OH)_{3}$ (s)+ $NaNO_{3}$

The precipitate that is formed is chromium hydroxide, $Cr(OH)_{3}$

When more NaOH is added, the precipitate reacts with it which then results in the formation of a soluble complex ion:

$Cr(OH)_{3}$(s) + $OH^{-}$(aq) → $Cr(OH)_{4} ^{-}$(aq)

$Cr(OH)_{4} ^{-}$ is soluble complex ion