A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I– concentration, which is 9.60×10-5 M, is 10,000 times less than that of the PO43– ion at 0.960 M . A solution containing the silver(I) ion is slowly added. Answer the questions below. Ksp of AgI is 8.30×10-17 and of Ag3PO4, 8.90×10-17.
Calculate the minimum Ag+ concentration required to cause precipitation of AgI.
Calculate the minimum Ag+ concentration required to cause precipitation of Ag3PO4.
¡) Dissolution equillibrium of AgI
AgI(s) <--------> Ag+(aq) + I-(aq)
Ksp = [Ag+] [I-]= 8.30×10^-17
given concentration of I- =9.60×10^-5M
Therefore,
[Ag+]×9.60×10^-5M = 8.30×10^-17M^2
[ Ag+ ] = 8.65×10^-13M
Therefore,
Minimum concentration of Ag+ required to cause precipitation of AgI is 8.65×10^-13M
ii) Dissociation equillibrium of Ag3PO4 is
Ag3PO4(s) --------> 3Ag+(aq) + PO43-
Ksp = [ Ag+ ]^3 [ PO43- ] = 8.90×10^-17
given concentration of PO43- = 0.960M
So,
[ Ag+ ]^3 × 0.960M = 8.90×10^-17M^4
[ Ag+ ]^3 = 9.27×10^-17M^3
[ Ag+ ] = 4.53 × 10^-6M
Therefore,
the minimum Ag+ concentration to cause precipitation of Ag3PO4 is 4.53×10^-6M
A basic solution contains the iodide and phosphate ions that are to be separated via selective...
Question: A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I– concentration, which is 9.10×10-5 M, is 10,000 times less than that of the PO43– ion at 0.910 M . A solution containing the silver(I) ion is slowly added. Answer the questions below. Ksp of AgI is 8.30×10-17 and of Ag3PO4, 8.90×10-17. Part 1. Calculate the minimum Ag+ concentration required to cause precipitation of AgI. Part 2. Calculate the minimum Ag+...
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