basic solution contains the iodide and phosphate ions that are to be separated v

Question

basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I– concentration, which is 8.60×10-5 M, is 10,000 times less than that of the PO43– ion at 0.860 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. (mol/L)

Part 2:

Calculate the minimum Ag+ concentration required to cause precipitation of Ag3PO4. (mol/L)

Explanation / Answer

answer :

[I-] =8.60*10-5  M

[po43-] =.860 M

Ksp of agI =8.30*10-17 ,

Ksp of ag3po4 =8.90*10-17

part 1 :

Ksp of agI =[ag+][I-]

8.30*10-17=[ag+]* 8.60*10-5

[Ag+]= (8.30* 10-17)/(8.60*10-5)= 0.965*10-12 M or 0.965 *10-5 mol/litre

part B

Ksp of Ag3PO4 =[Ag+]3[PO43-]

8.90*10-17=[Ag+]3 * 0.860

[Ag+]3 =(8.90 *10-17) /(0.860)=10.348 *10-17

[Ag+]=4.69 *10-6M or 4.69 *10-6 mol/litre

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