copper(I) ion forms a chloride salt that has Ksp = 1.2×10^-6

copper(I) ion forms a chloride salt that has Ksp = 1.2×10^-6

 

 

The copper(I) ion forms a chloride salt that has Ksp = 1.2×10^-6. Copper(I) also forms a complex ion with Cl^-:

Cu^+(aq) + 2Cl^-(aq) ⇌ CuCl2^-(aq) ... Kf = 8.7×10^4

(a) Calculate the solubility of copper(I) chloride in pure water. (Ignore CuCl2^- formation for part a).

(b) Calculate the solubility of copper(I) chloride in 0.10 M NaCl.

 

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(a)

난용성염의 용해도

CuCl(s) ⇌ Cu^+(aq) + Cl^-(aq)

 

 

Ksp = [Cu^+] [Cl^-]

1.2×10^(-6) = (x) (x)

x = (1.2×10^(-6))^(1/2)

= 0.0011 M

= 1.1×10^(-3) M

( 참고 https://ywpop.tistory.com/3141 )

 

x = [Cu^+] = [Cl^-] = [CuCl]

 

 

 

(b)

두 반응식을 더하면,

consider both Ksp reaction and Kf reaction at the same time

 

CuCl(s) ⇌ Cu^+(aq) + Cl^-(aq) ... Ksp = 1.2×10^(-6)

Cu^+(aq) + 2Cl^-(aq) ⇌ CuCl2^-(aq) ... Kf = 8.7×10^4

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CuCl(s) + Cl^-(aq) ⇌ CuCl2^-(aq) ... K = 0.10

 

K = Ksp × Kf = (1.2×10^(-6)) × (8.7×10^4) = 0.1044 ≒ 0.10

( 참고 https://ywpop.tistory.com/4154 )

 

 

ICE 도표를 작성하면,

[Cl^-] = [NaCl] = 0.10 M

 

.........CuCl(s) + Cl^-(aq) ⇌ CuCl2^-(aq)

초기(M) .......... 0.10 ...... 0

변화(M) .......... -x ........ +x

평형(M) .......... 0.10-x .... x

 

 

K = [CuCl2^-] / [Cl^-]

0.10 = (x) / (0.10-x)

 

x = (0.10)^2 – 0.10x

x = (0.10)^2 / 1.10

= 0.0091 M

= 9.1×10^(-3) M

 

x = [CuCl2^-] = [CuCl]

 

 

 

[키워드] 평형상수의 이론적 계산 기준문서, Ksp와 Kf 기준문서, CuCl의 몰용해도 기준문서, 염화구리(I)의 용해도 기준문서, 염화구리(I)의 몰용해도 기준문서