The addition of dilute NaOH to Cr\(^{3+}\) salt solution will give :
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Remember the amphoteric nature of certain metal hydroxides like Al(OH)\(_3\), Zn(OH)\(_2\), Pb(OH)\(_2\), and Cr(OH)\(_3\). They precipitate with a small amount of base and redissolve in excess strong base. The term "dilute" usually implies precipitation is the final step.
Step 1: Understanding the Question:
We need to predict the product when a dilute solution of sodium hydroxide (NaOH) is added to a solution containing chromium(III) ions (Cr\(^{3+}\)). Step 2: Detailed Explanation:
Chromium(III) hydroxide, Cr(OH)\(_3\), is amphoteric, meaning it can react with both acids and bases.
1. Reaction with a limited amount of base: When a dilute or limited amount of a base like NaOH is added to a Cr\(^{3+}\) solution, the hydroxide ions react with the chromium ions to form a gelatinous, greyish-green precipitate of chromium(III) hydroxide.
The chemical equation for this precipitation reaction is:
\[ \text{Cr}^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Cr(OH)}_3(s) \]
2. Reaction with excess strong base: If an excess of a strong base (like concentrated NaOH) were added, the amphoteric precipitate would redissolve to form a soluble, green-colored complex ion, tetrahydroxochromate(III), [Cr(OH)\(_4\)]\(^-\).
\[ \text{Cr(OH)}_3(s) + \text{OH}^-(aq) \rightarrow [\text{Cr(OH)}_4]^-(aq) \]
Since the question specifies the addition of dilute NaOH, this implies a limited amount of base is used, leading to the formation of the precipitate, not its redissolution. The formula Cr\(_2\)O\(_3\)(H\(_2\)O)\(_n\) represents hydrated chromium(III) oxide, which is essentially the same as Cr(OH)\(_3\), but Cr(OH)\(_3\) is the conventional way to write the formula for the hydroxide precipitate. Step 3: Final Answer:
The addition of dilute NaOH to a Cr\(^{3+}\) salt solution will give a precipitate of Cr(OH)\(_3\).
