Question

Suppose that a 10-mL 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?

Answer 1

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^-.

Answer 2

1e-7

Answer 3

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)=1.77E-10
1.77E-10=[.oo3M Ag(+)][Cl^(-)]
solve for Cl^(-)

Cl^(-)=5.9E-8M
5.9E-8mol/L*.01L in tank * mm 35.453 g/mol=2*10^-8 g