calculate the molar solubility of AgI in a 1.0 M NH3 solution Solution AgI = > A

Question

calculate the molar solubility of AgI in a 1.0 M NH3 solution

Explanation / Answer

AgI = > Ag+ + I- Ksp = [Ag+ ] [ I-] = 8.3 * 10 ^ - 17 Ag+ + 2NH3 Ag(NH3)2+ Kf = [Ag(NH3)2+] / [Ag+] [ NH3-] ^2 = 1.7 * 10 ^ 7 almost every Ag- ion formed from AgI dissolution is transformed into the complex. ( Kf is very large and favours the RHS) For each Ag + that dissolves 1 I - ion is released so the [I-] ~ [Ag(NH3)2+] = c Combining the equation and solving for [Ag+} c^2 = Ksp * Kf * [NH3] ^2 c^2 = 8.3 * 10 ^-17 * 1.70 ^7 * (2.4-2c) ^2 c = 9 * 10 ^ -5 moles per liter [I-] which results from the dissolution of the AgI = 9 * 10 -5 moles/ liter solubility of AgI in 2.4 M NH3 is 9 * 10 ^ -5 molar in pure water it would have been [Ag+ ] [ I-] = 8.3 * 10 ^ - 17 [I-] = 9.1* 10 -7 or 100 times less soluble

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