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

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

Step-by-step 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