0.10 mol AgBr 3.0 M NH3 1.0 L Ksp 5.0×10^(-13) Ag(NH3)2^+

Will 0.10 mol of AgBr completely dissolve in 1.0 L of 3.0 M NH3?

The Ksp value for AgBr(s) is 5.0×10^(-13),

and the overall formation constant for

the complex ion Ag(NH3)2^+ is 1.7×10^7, that is,

Ag^+(aq) + 2NH3(aq) ⇌ Ag(NH3)2^+(aq) ... K = 1.7×10^7

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.. AgBr(s) ⇌ Ag^+(aq) + Br^-(aq) ... Ksp = 5.0×10^(-13)

+) Ag^+(aq) + 2NH3(aq) ⇌ Ag(NH3)2^+(aq) ... Kf = 1.7×10^7

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.. AgBr(s) + 2NH3(aq) ⇌ Ag(NH3)2^+(aq) + Br^-(aq) ... Kc = ?

Kc = (5.0×10^(-13)) × (1.7×10^7) = 8.5×10^(-6)

... AgBr(s) + 2NH3(aq) ⇌ Ag(NH3)2^+(aq) + Br^-(aq)

평형에서 ....... 3.0–2x ............... x ................................................ x

Kc = [Ag(NH3)2^+] [Br^-] / [NH3]^2

8.5×10^(-6) = x^2 / (3.0–2x)^2

(x / (3.0–2x))^2 = 8.5×10^(-6)

x / (3.0–2x) = (8.5×10^(-6))^(1/2) = 0.002915476

x = 0.002915476 × (3.0–2x)

x = 0.008746428 – 0.005830952x

1.005830952x = 0.008746428

x = 0.008746428 / 1.005830952

= 0.0087 M = [Br^-] = [AgBr]

---> 3.0 M NH3에서 AgBr의 몰용해도

0.10 mol of AgBr completely dissolve in 1.0 L

---> 0.10 mol / 1.0 L = 0.10 M AgBr

0.10 M > 0.0087 M 이므로,

즉, [AgBr]는 몰용해도보다 높기 때문에,

AgBr은 침전된다.

( 완전히 용해되지 않는다. )

[ 관련 예제 https://ywpop.tistory.com/8224 ]

1.0 M NH3에서 AgBr의 몰용해도

[키워드] AgBr의 몰용해도 기준, AgBr의 Ksp 기준, Ag(NH3)2^+의 Kf 기준