0.50 M NH4Cl에서 Mg(OH)2의 용해도

0.50 M NH4Cl에서 Mg(OH)2의 용해도

 

 

Calculate the solubility of Mg(OH)2 in 0.50 M NH4Cl.

Kb for NH3 = 1.8×10^(-5)

Ksp for Mg(OH)2 = 1.8×10^(-11)

 

 

 

Mg(OH)2(s) ⇌ Mg^2+(aq) + 2OH^-(aq)

Ksp = [Mg^2+] [OH^-]^2

 

 

 

The solubility of Mg(OH)2 is increased by addition of NH4^+ ion.

( 참고: NH4Cl은 산성염 https://ywpop.tistory.com/12986 )

 

2NH4^+(aq) + Mg(OH)2(s) ⇌ 2NH3(aq) + 2H2O(l) + Mg^2+(aq)

Kc = [NH3]^2 [Mg^2+] / [NH4^+]^2

 

 

 

NH3(aq) + H2O(l) ⇌ NH4^+(aq) + OH^-(aq)

Kb = [NH4^+] [OH^-] / [NH3]

 

 

 

 

Kc × Kb^2

= ([NH3]^2 [Mg^2+] / [NH4^+]^2) × ([NH4^+]^2 [OH^-]^2 / [NH3]^2)

= [Mg^2+] [OH^-]^2 = Ksp

 

 

 

Kc = Ksp / Kb^2

= (1.8×10^(-11)) / (1.8×10^(-5))^2

= 0.05556

 

 

 

 

............. 2NH4^+(aq) + Mg(OH)2(s) ⇌ 2NH3(aq) + 2H2O(l) + Mg^2+(aq)

초기(M) . 0.5 ......................................... 0 ................................. 0

변화(M) . –2x ........................................ +2x .............................. +x

평형(M) . 0.5–2x ................................... 2x ................................ x

 

Kc = (2x)^2 (x) / (0.5-2x)^2

 

 

 

0.5–2x ≒ 0.5 라고 근사처리하면,

( 근사처리할 수 없으나,

x^3 계산이 곤란하므로,

대략적인 값을 계산해보면, )

 

4x^3 / 0.5^2 = 0.05556

 

x = [0.05556 × 0.5^2 / 4]^(1/3)

= 0.1514 M

≒ 0.15 M = [Mg^2+]

---> Mg(OH)2의 몰용해도

 

 

 

 

[참고] Ksp for Mg(OH)2 = 1.0×10^(-11) 인 경우

Kc = Ksp / Kb^2

= (1.0×10^(-11)) / (1.8×10^(-5))^2

= 0.03086

 

 

x = [0.03086 × 0.5^2 / 4]^(1/3)

= 0.1245 M

≒ 0.12 M

 

 

 

 

[키워드] Mg(OH)2의 몰용해도 기준