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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?

  • Chemistry -

    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^-.

  • Chemistry -


  • Chemistry -

    AgNO3= 2E^-4L*0.15mol/L
    AgNO3/(.01L)=mol/L AgNO3=mol/L Ag^(+)=.003M

    Find Ksp(AgCl) in solubility constant table
    1.77E-10=[.oo3M Ag(+)][Cl^(-)]
    solve for Cl^(-)

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

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