Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\text{Ag}^+$ and $\text{Pb}^{2+}$ at a concentration of $0.10 \text{ M}$ . Aqueous $\text{HCl}$ is added to this solution until the $\text{Cl}^-$ concentration is $0.10 \text{ M}$ . What will the concentration of $\text{Ag}^+$ and $\text{Pb}^{2+}$ at equilibrium? ( $K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$ , $K_{sp}$ for $\text{PbCl}_2 = 1.7 \times 10^{-5}$ )

$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-4} \text{ M}$

$[\text{Ag}^+] = 1.8 \times 10^{-7} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-6} \text{ M}$

$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 8.5 \times 10^{-5} \text{ M}$

$[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}$

Step-by-Step Solution

Given that the equilibrium concentration of $\text{Cl}^-$ is $[\text{Cl}^-] = 0.10 \text{ M}$ .

For $\text{AgCl}$ , the solubility product is given by: $K_{sp}(\text{AgCl}) = [\text{Ag}^+][\text{Cl}^-]$ $1.8 \times 10^{-10} = [\text{Ag}^+] \times 0.10$ $[\text{Ag}^+] = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \text{ M}$

For $\text{PbCl}_2$ , the solubility product is given by: $K_{sp}(\text{PbCl}_2) = [\text{Pb}^{2+}][\text{Cl}^-]^2$ $1.7 \times 10^{-5} = [\text{Pb}^{2+}] \times (0.10)^2$ $1.7 \times 10^{-5} = [\text{Pb}^{2+}] \times 0.01$ $[\text{Pb}^{2+}] = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3} \text{ M}$

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