In qualitative analysis, group III cations (such as \( \text{Fe}^{3+}, \text{Cr}^{3+}, \text{Al}^{3+} \)) are precipitated as hydroxides by adding ammonium hydroxide \( \text{NH}_4\text{OH} \).
Ammonium chloride \( \text{NH}_4\text{Cl} \) is added before ammonium hydroxide to control the concentration of \( \text{OH}^- \) ions. This is achieved through the common ion effect:
\[ \text{NH}_4\text{OH} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
Adding \( \text{NH}_4\text{Cl} \) increases the concentration of \( \text{NH}_4^+ \) ions, which shifts the equilibrium to the left, decreasing the concentration of \( \text{OH}^- \) ions.
By reducing \( \text{OH}^- \) concentration, we avoid the formation of precipitates from cations of higher groups (such as Group IV and V cations), ensuring selective precipitation of Group III cations only.
The addition of ammonium chloride decreases the concentration of \( \text{OH}^- \) ions through the common ion effect, which corresponds to Option (2).
Match List I with List II:
Choose the correct answer from the options given below:
The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ____. Use: Atomic mass of N (in amu) = 14
The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions is:
List-I | List-II |
---|---|
(P) Passing H2S in the presence of NH4OH | (1) Cu2+ |
(Q) (NH4)2CO3 in the presence of NH4OH | (2) Al3+ |
(R) NH4OH in the presence of NH4Cl | (3) Mn2+ |
(S) Passing H2S in the presence of dilute HCl | (4) Ba2+ (5) Mg2+ |
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: