PbI2 200.0 mL 240 mg PbI2 Ksp

PbI2 200.0 mL 240 mg PbI2 Ksp

On a hot day, a 200.0 mL sample

of a saturated solution of PbI2

was allowed to evaporate until dry.

If 240 mg of solid PbI2 was collected

after evaporation was complete,

calculate the Ksp value for PbI2 on this hot day.

> 200.0 mL = 200.0/1000 L

> 240 mg = 240/1000 g

PbI2의 몰질량 = 461.01 g/mol 이므로,

(240/1000 g) / (461.01 g/mol)

= 0.00052 mol PbI2

( 참고 https://ywpop.tistory.com/7738 )

[PbI2] = 0.00052 mol / (200.0/1000 L)

= 0.0026 M

---> PbI2의 몰용해도(s)

PbI2(s) ⇌ Pb^2+(aq) + 2I^-(aq)

Ksp = [Pb^2+] [I^-]^2

= (s) (2s)^2

= 4s^3

( 참고 https://ywpop.tistory.com/8434 )

= 4 × (0.0026)^3

= 7.0×10^(-8)

답: 7.0×10^(-8)

[키워드] PbI2의 몰용해도 기준, PbI2의 Ksp 기준