1.0×10^(-2) M Pb(NO3)2 100.0 mL 1.0×10^(-3) M NaF 100.0 mL

1.0×10^(-2) M Pb(NO3)2 100.0 mL 1.0×10^(-3) M NaF 100.0 mL

PbF2(s)의 Ksp = 4×10^(-8)

 

 

A solution is prepared by mix

100 mL of 1.0×10^(-2) M Pb(NO3)2 and

100.0 mL of 1.0×10^(-3) M NaF.

Will PbF2(s) (Ksp = 4×10^(-8)) precipitate?

 

---------------------------------------------------

 

> Pb(NO3)2의 몰수 = (1.0×10^(-2) mmol/mL) (100.0 mL) = 1.0 mmol

> NaF의 몰수 = (1.0×10^(-3)) (100.0) = 0.10 mmol

( 참고 https://ywpop.tistory.com/7787 )

 

 

 

침전 반응

Pb(NO3)2(aq) + 2NaF(aq) → PbF2(s) + 2NaNO3(aq) 또는

Pb^2+(aq) + 2F^-(aq) → PbF2(s)

 

 

 

PbF2(s) ⇌ Pb^2+(aq) + 2F^-(aq)

Q = [Pb^2+][F^-]^2

 

 

 

혼합 후,

[Pb^2+] = 1.0 mmol / 200.0 mL = 0.0050 M

[F^-] = 0.10 mmol / 200.0 mL = 0.00050 M

 

 

 

Q = (0.0050) (0.00050)^2 = 1.3×10^(-9)

 

 

 

답: Q < Ksp 이므로,

( 참고 https://ywpop.tistory.com/12527 )

침전되지 않음.

 

 

 

 

[키워드] Q와 Ksp 기준문서, Ksp와 Q 기준문서