JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

For a sparingly soluble salt \(\mathrm{AB}_2\), the equilibrium concentrations of \(\mathrm{A}^{2+}\) ions and \(\mathrm{B}^{-}\)ions are \(1.2 \times 10^{-4} \mathrm{M}\) and \(0.24 \times 10^{-3} \mathrm{M}\), respectively. The solubility product of \(\mathrm{AB}_2\) is _______.

A \(0.069 \times 10^{-12}\)
B \(6.91 \times 10^{-12}\)
C \(0.276 \times 10^{-12}\)
D \(27.65 \times 10^{-12}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(6.91 \times 10^{-12}\)

Step-by-step Solution

Detailed explanation

\( \mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{+2}+2 \mathrm{~B}_{(\mathrm{aq})}^{-} \) \( \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+2}\right]\left[\mathrm{B}^{-}\right]^2 \) \(=1.2 \times 10^{-4} \times\left(2.4 \times 10^{-4}\right)^2 \)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List\(-I\) with List\(-II :\)

List\(-I\) (Chemical Reaction)	List\(-II\) (Reagent used)
\((a)\) \(\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)	\((i)\) \(\mathrm{CH}_{3} \mathrm{MgBr} / \mathrm{H}_{3} \mathrm{O}^{+}\) \((1 .\) equivalent \()\)
\((b)\) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3} \rightarrow \mathrm{CH}_{3} \mathrm{CHO}\)	\((ii)\) \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{H}_{2} \mathrm{O}\)
\((c)\) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N} \rightarrow \mathrm{CH}_{3} \mathrm{CHO}\)	\((iii)\) \(\mathrm{DIBAL}-\mathrm{H} / \mathrm{H}_{2} \mathrm{O}\)
\((d)\) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N} \rightarrow \mathrm{CH}_{3}CO\mathrm{CH}_{3}\)	\((iv)\) \(\mathrm{SnCl}_{2}, \mathrm{HCl} / \mathrm{H}_{2} \mathrm{O}\)

Choose the most appropriate match :

JEE Mains 2021 Medium

The reaction \(2 NO + Br _2 \rightarrow 2 NOBr\) takes places through the mechanism given below : \(NO + Br _2 \Leftrightarrow NOBr _2 \text { (fast) }\) \(NOBr _2+ NO \rightarrow 2 NOBr \text { (slow) }\) The overall order of the reaction is \(.....\).

JEE Mains 2023 Medium

Match List \(I\) with List \(II\).

List \(-I\)(Element)	List \(-II\) (Electronic Configuration)
\(A.\) \(N\)	\(I.\) \([\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^5\)
\(B.\) \(S\)	\(II.\) \([\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4\)
\(C.\) \(Br\)	\(III.\) \([\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^3\)
\(D.\) \(Kr\)	\(IV.\) \([\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^6\)

Choose the correct answer from the options given below:

JEE Mains 2024 Medium

Explore more questions on app

From JEE Mains

Explore more questions on app