If the solubility product of PbS is 8 x 10^(-28), then the solubility of PbS in pure water at 298 K is x x 10^(-16) mol L^(-1). The value of x is __________. (Nearest Integer) [Given : sqrt2 = 1.41]

K_ sp= S^2 S=sqrtK_s p=sqrt8 x 10^-28=2 sqrt(2) x 10^-14 =2.82 x 10^-14 =282 x 10^-16 Therefore Ans. 282

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Chemistry

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Chemistry

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]

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Explanation

$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}\\$ $\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}\\$ $=2.82 \times 10^{-14}\\$ $=282 \times 10^{-16}\\$ $\therefore$ Ans. 282