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 __________. (Nearest Integer)
[Given : $\sqrt2$ = 1.41]
Answer (integer)
282
Solution
$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}$
<br/><br/>
$\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}$
<br/><br/>
$=2.82 \times 10^{-14}$
<br/><br/>
$=282 \times 10^{-16}$
<br/><br/>$\therefore$ Ans. 282
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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