If the solubility product of PbS is 8 × 10-28, then the solubility of PbS in pure water at 298 K is x × 10-16 mol L -1. The value of x is . (Nearest Integer) [Given √2=1.41 ]

Question

If the solubility product of PbS is 8 × 10-28, then the solubility of PbS in pure water at 298 K is x × 10-16 mol L -1. The value of x is . (Nearest Integer) [Given √2=1.41 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K sp = S 2 S =√ K sp =√8 × 10-28=2 √2 × 10-14 =2.82 × 10-14 =282 × 10-16 Ans. =282

Q. If the solubility product of $P b S$ is $8 \times 1 0^{- 28}$ , then the solubility of $P b S$ in pure water at $298 K$ is $x \times 1 0^{- 16} m o l L^{- 1}$ . The value of $x$ is ___ . (Nearest Integer)
[Given $2 = 1.41$ ]

$K_{s p} = S^{2}$
$S = K_{s p} = 8 \times 1 0^{- 28} = 22 \times 1 0^{- 14}$
$= 2.82 \times 1 0^{- 14}$
$= 282 \times 1 0^{- 16}$
Ans. = $282$

Q. If the solubility product of PbS is 8×10−28, then the solubility of PbS in pure water at 298K is x×10−16molL−1. The value of x is ___ . (Nearest Integer) [Given 2​=1.41 ]

Ksp​=S2 S=Ksp​​=8×10−28​=22​×10−14 =2.82×10−14 =282×10−16 Ans. =282

Q. If the solubility product of $P b S$ is $8 \times 1 0^{- 28}$ , then the solubility of $P b S$ in pure water at $298 K$ is $x \times 1 0^{- 16} m o l L^{- 1}$ . The value of $x$ is ___ . (Nearest Integer)
[Given $2 = 1.41$ ]

$K_{s p} = S^{2}$
$S = K_{s p} = 8 \times 1 0^{- 28} = 22 \times 1 0^{- 14}$
$= 2.82 \times 1 0^{- 14}$
$= 282 \times 1 0^{- 16}$
Ans. = $282$