JEE Advanced · Chemistry · 8. Ionic Equilibrium

In $1 \mathrm{~L}$ saturated solution of $\mathrm{AgCl}\left[K_{\mathrm{sp}} \mathrm{AgCl}=1.6 \times 10^{-10}\right], 0.1$ mole of $\mathrm{CuCl}\left[K_{\mathrm{sp}}(\mathrm{CuCl})=1.0 \times 10^{-6}\right]$ is added. The resultant concentration of $\mathrm{Ag}^{+}$in the solution is $1.6 \times 10^{-x}$. The value of $x$ is

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Answer & Solution

Correct Answer

(C) 7

Step-by-step Solution

Detailed explanation

It is a case of simultaneous solubility of salts with a common ion. Here solubility product of $\mathrm{CuCl}$ is much greater than that of $\mathrm{AgCl}$, it can be assumed that $\mathrm{Cl}^{-}$in solution comes mainly from $\mathrm{CuCl}$.
$\Rightarrow \left[\mathrm{Cl}^{-}\right]=\sqrt{K_{s p}(\mathrm{CuCl})}=10^{-3} \mathrm{M}$
Now for $\mathrm{AgCl}: K_{s p}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=$ $\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \Rightarrow\left[\mathrm{Ag}^{+}\right]=1.6 \times 10^{-7}$

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.

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From JEE Advanced

A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v}$ =0	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = - B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = - B_{0} \hat{z}$

In which case would the particle move in a straight line along the negative direction of y axis

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Atomic number	Ionization Enthalpy (kJ/mol)
	$I_{1}$	$I_{2}$	$I_{3}$
$n$	$1681$	$3374$	$6050$
$n + 1$	$2081$	$3952$	$6122$
$n + 2$	$496$	$4562$	$6910$
$n + 3$	$738$	$1451$	$7733$