Chemistry
posted by Josh on .
Suppose that a 10mL sample of a solution is to be tested for Cl ion by addition of 1 drop (0.2mL ) of 0.15 M AgNO3 .
What is the minimum number of grams of Cl that must be present in order for AgCl(s) to form?

AgCl ==> Ag^+ + Cl^
Ksp = (Ag^+)(Cl^)
You know 0.2 mL x 0.15 M AgNO3 will be moles of M x L. That divided by 10 mL (convert to L) will be the concn of Ag^+. Solve for (Cl^) which will be in moles/L, convert that to moles/10 mL and convert moles Cl^ to grams Cl^. 
1e7

AgCl>Ag^(+)+Cl^()
AgNO3= 2E^4L*0.15mol/L
AgNO3/(.01L)=mol/L AgNO3=mol/L Ag^(+)=.003M
Find Ksp(AgCl) in solubility constant table
Ksp(AgCl)=1.77E10
1.77E10=[.oo3M Ag(+)][Cl^()]
solve for Cl^()
Cl^()=5.9E8M
5.9E8mol/L*.01L in tank * mm 35.453 g/mol=2*10^8 g