Question

**Calculate the molar solubility of AgBr in a 2.2**
*M* **NH**_{3}
**solution**

Answer #1

AgBr = > Ag+ + Br-

Ksp = [Ag+ ] [ Br-] = 7.7 * 10 ^ - 13

Ag+ + 2NH3 <=> Ag(NH3)2+

Kf = [Ag(NH3)2+] / [Ag+] [ NH3-] ^2 = 1.7 * 10 ^ 7

almost every Ag- ion formed from AgBr dissolution is transformed
into the complex. ( Kf is very large and favours the RHS)

For each Ag + that dissolves 1 Br- ion is released

so the [Br-] ~ [Ag(NH3)2+] = c

Combining the equation and solving for [Ag+}

c^2 = Ksp * Kf * [NH3] ^2

c^2 = 7.7 * 10 ^-13 * 1.70 *10 ^7 * (2.2-2c) ^2

c = 7.9 * 10 ^ -3moles per liter

**[Br-] which results from the dissolution of the AgBr = 7.9
* 10 -3 moles/ liter
solubility of AgBr in 2.2 M NH3 is 7.9 * 10 ^ -3 molar**

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