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Homework Help: Solubility Product

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the minimum quantity of ammonia which must be added to 1L of a solution in order to dissolve 0.1 mole silver chloride by forming [Ag(NH3)2]+.
    Given Ksp- 10^-10 and Kf- 10^8


    2. Relevant equations



    3. The attempt at a solution

    I used the equation (which I derived) s=c^2/(Ksp)(Kf) (assuming almost all amount of AgCl converts to complex and some other approximations)
    where s is the solubility of AgCl and c is the concentration of NH3 (both in mol/litre)

    I solved and got c as 1 mole.
    But answer given is 1.2 mole.

    Are my assumptions wrong, or there is some other thing?
     
  2. jcsd
  3. Sep 23, 2010 #2

    Borek

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    Staff: Mentor

    Show your derivation.
     
  4. Sep 23, 2010 #3
    Let s be the solubility of AgCl in cM NH3. Let x mole/L be the amount of salt forming complex

    At equilibrium.
    Ag+ = s-x
    ksp=(s-x)(s)
    kf= x/(s-x)(c-2x)^2
    Since kf is very high, x approaches s. Thus s-x would be very small. Let this value be y. Since x is small, 2x can be ignored w.r.t c.

    i.e. c-2x = c
    ksp=ys
    kf=s^2/ksp(c)^2
    i.e.
    s=c x (ksp x kf)^1/2

    Sorry, I wrote the earlier expression wrong.
     
  5. Sep 23, 2010 #4

    Borek

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    Staff: Mentor

    You want x to be around 0.1M. It is not small.
     
  6. Sep 24, 2010 #5
    Oops. Then that will make the calculations very lengthy.
    Is there any method by which we solve this way and then add 0.2 mole by giving some reason?
     
  7. Sep 24, 2010 #6

    Borek

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    Staff: Mentor

    What is [Cl-]?

    What is maximum possible [Ag+]?

    What is complex concentration?

    Calculate ammonia concentration that fits.

    --
     
  8. Sep 24, 2010 #7
    max [CL-] is 0.1
    max [Ag+] is 10^-9

    How to calculate ammonia concentratrion that fits?
     
  9. Sep 24, 2010 #8

    Borek

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    Staff: Mentor

    What is complex concentration?

    Look at Kf.
     
  10. Sep 24, 2010 #9
    Complex concentration will be practically 0.1.
    From this
    [NH3]^2 = 0.1/(10^-9)(10^8)
    I got [NH3] as 1M
    not 1.2.
     
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