Silver nitrate is slowly added to a solution containing 0.020 M Cl^- and 0.020 M
ID: 1054596 • Letter: S
Question
Silver nitrate is slowly added to a solution containing 0.020 M Cl^- and 0.020 M Br^-. Calculate the concentration of Ag+ ions (in mole/L) required to initiate the precipitation of AgCl AND what is the concentration of Br- remaining just before AgCl begins to precipitate, KspAgBr = 7.7 times 10^-13, Ksp AgCl = 1.6 times 10^-10 The concentration of a calcium ion, Ca^2+, in a blood plasma is 0.0025 M. If the concentration of oxalate ion, C_2CO_4^2-, is 1.0 times 10^-7 M, do you expect calcium oxalate, CaC_2O_4 to precipitate ? K_spCaC_2O_4 = 2.3 times 10^-9 What is the standard emf, E_cell, for the following galvanic cell Cr(s) I Cr^3+ (aq)(0.10 M) II KCl sat.) II Hg_2^2+ (aq) (0.01 M) I 2Hg(l)Explanation / Answer
4-
Ksp for AgCl is 1.6 x10^10
Ksp for AgBr is 7.7 x10^13
Since Ksp for AgBr is less than Ksp for AgCl, so AgBr is less soluble. It will precipitate first.
The solubility equilibria for AgBr
AgBr <-----> Ag+(aq) + Br-(aq)
[Br-] = 0.020M
Ksp = [Ag+] x [Br-]
[Ag+] = 7.7 x 10^-13/0.020 = 3.9*10^-11 M
The concentrations of Ag+ necessary to initiate the precipitation is > 3.9*10^-11 M
The concentrations of Ag+ (in mole/L) necessary to initiate the precipitation of AgCl
Given-
The solubility equilibria for AgCl
AgCl <---> Ag+(aq) + Cl-(aq)
Ksp for AgCl is 1.6 x10^10
[Cl-] = 0.020M
Ksp = [Ag+] x [Cl-]
[Ag+] =1.6 x 10^-10/0.020 = 8.0 x 10^-9 M
The concentrations of Ag+ necessary to initiate the precipitation is > 8.0 x 10^-9 M
The Cl- will be precipitate only when [Ag+] reached 8.0 x 10^-9 M
Therefore concentration of remaining Br- = Ksp for AgBr /[Ag+]
= 7.7 x 10^-13 /8.0 x 10^-9 M
= 9.63 x 10^-5M
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