If solubility product of PbS is 8 × 10^{-28} , then solubility of PbS in pure water at 298 K is x ?

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6-2.Equilibrium-II (Ionic Equilibrium)

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If the solubility product of $PbS$ is $8 \times 10^{-28}$, then the solubility of $PbS$ in pure water at $298\; K$ is $x \times 10^{-16}\; mol\; L ^{-1}$. The value of $x$ is $\dots$.

[Given $\sqrt2 = 1.41$]

A

$281$

B

$282$

C

$283$

D

$284$

(JEE MAIN-2022)

Solution

$K _{ sp }= S ^{2}$

$S =\sqrt{ K _{ sp }}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$

$=2.82 \times 10^{-14}$

$=282 \times 10^{-16}$

Ans. $=282$

Standard 11

Chemistry

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