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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.

Respuesta :

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:

[tex]Cr(NO)_{3}[/tex] +3 NaOH → [tex]Cr(OH)_{3}[/tex] (s)+ [tex]NaNO_{3}[/tex]

The precipitate that is formed is chromium hydroxide, [tex]Cr(OH)_{3}[/tex]

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

[tex]Cr(OH)_{3}[/tex](s) + [tex]OH^{-}[/tex](aq) → [tex]Cr(OH)_{4} ^{-}[/tex](aq)

[tex]Cr(OH)_{4} ^{-}[/tex] is soluble complex ion

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