Could anyone help me with these two questions? Please explain :) thank you!! Qua
ID: 1007951 • Letter: C
Question
Could anyone help me with these two questions? Please explain :) thank you!! Quartative Analysis: Cations, Anions, and Complex lons imes be brought into solution by the formation of a soluble complex can be dissolved in ammonia solutions by the formation of the ion. The ammonia competes with the CI ion for reaction with Ag 4) Insoluble salts can sometimes be brought into solution by the ion. For example, AgCI(s) soluble Ag(NH)2 complex ion. The ammonia competes w as follows: AgCI(s) Ag+(aq) + Cl-(aq) Ag+(aq) + 2NH3(aq) Ag(NH3)20w) K,,-[Ag+][C] = 1.8 × 10'10 Amation = Sum: AgCI(s) + 2NFldag) Ag(NH):Uaq) + Cl-(aq) = 2.9x10 Assuming all the CI comes from the dissociation of AgCI, then [Ag(NH)- [CT J, and [Ag(NH, ),'/>[NHJ. x 2.9x10" or [Ag(NH),, 1-1NHJxV2.9×10-3 What is the maximum [Ag(NHs)2'1 that can be obtained by dissolving AgCl in 1.5 M ammonia? For another silver salt, AgBrO, K,-5.4 × 10-5. what is Krt for the reaction AgBrd) + 2NH3 Ag(NH),+ + Bro? What is the maximum [Ag(NHs)21 that can be obtained by dissolving AgBrO, in 4.1 NM 5) ammonia?Explanation / Answer
Q.4: The final equation we obtained is [Ag(NH3)2+] = [NH3] x squareroot(2.9x10-3)
=> [Ag(NH3)2+] = [NH3] x 0.053852
Given [NH3] = 1.5 M
=> [Ag(NH3)2+] = 1.5 M x 0.053852
=> [Ag(NH3)2+] = 0.0808 M (answer)
Hence the maximum concentration of [Ag(NH3)2+] that can be obtained is 0.0808 M.
Q.5: AgBrO3(s) <--------> Ag+(aq) + BrO3-(aq) : Ksp = 5.4x10-5 ---- (1)
Ag+(aq) + 2NH3(aq) <--------> Ag(NH3)2+(aq) : Kformation = 1.6x107 -----(2)
By adding equation (1) and (2) we will get
AgBrO3(s) + 2NH3(aq) <--------> Ag(NH3)2+(aq) + BrO3-(aq) : Knet
Knet = Ksp x Kformation
=> Knet = 5.4x10-5 x 1.6x107
=> Knet = 8.64x102 (answer)
Also
[Ag(NH3)2+] = [NH3] x squareroot(Knet)
=> [Ag(NH3)2+] = [NH3] x squareroot(8.64x102) = [NH3] x 29.40
Given [NH3] = 4.1 M
=> [Ag(NH3)2+] = 4.1 M x 29.40 = 120.5 M (answer)