PbI2의 몰용해도. Ksp = 7.9×10^(-9)

PbI2의 몰용해도. Ksp = 7.9×10^(-9)

 

 

PbI2의 용해도를 구하시오.

PbI2의 Ksp = 7.9×10^(-9)

 

 

The solubility product of PbI2 is 7.9×10^(-9).

What is the molar solubility of PbI2 in distilled water?

 

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PbI2(s) ⇌ Pb^2+(aq) + 2I^-(aq)

 

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

= (s) (2s)^2

= 4s^3

( 참고 https://ywpop.tistory.com/8434 )

 

 

 

몰용해도, s를 계산하면,

s = [Ksp / 4]^(1/3)

= [(7.9×10^(-9)) / 4]^(1/3)

= 0.00125465 M

= 1.3×10^(-3) M

 

 

 

 

[참고]

PbI2의 몰질량 = 461.01 g/mol 이므로,

(0.00125465 mol/L) × (461.01 g/mol)

= 0.578406 g/L

---> PbI2는 물 1 L당 0.5784 g 용해 가능.

 

 

 

 

> [Pb^2+] = 0.00125465 M

= 1.3×10^(-3) M

 

> [Cl^-] = 2 × 0.00125465 = 0.0025093 M

= 2.5×10^(-3) M