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Homework Help: How many times more soluble. HELP

  1. Dec 19, 2005 #1
    How many times more soluble........ ASAP HELP

    Hii, I did a lab and need help with this question it says

    How many times more soluble is lead 2 chloride in distilled water than in 0.100 mol/L solution of sodium chloride?

    The ksp for lead 2 chloride is = 1.2*10^-5

    I dont know what other information is needed to solve this question or how to figure this out.
    ksp=[Pb] [Cl]^2

  2. jcsd
  3. Dec 19, 2005 #2
    There is no additional information required.

    The reaction is
    [tex]PbCl_2 \, \Leftrightarrow Pb^{2+} + 2Cl^-[/tex]

    In distilled water, there is no Pb2+ or Cl- present so [Pb2+] at equilibrium is that formed by the dissociation of PbCl2 only.

    Table-wise, if we let the concentration of Pb2+ formed by s:

    [tex] \begin{array}{ c|ccc|}{}&{PbCl_2 \, \Leftrightarrow}&{Pb^{2+}}&{2Cl^-}\\
    \hline{\text{Initial conc.}}&{}&0&0\\
    \text{Change in conc.}}&{}&{+s}&{+2s}\\
    \text{Conc. at equi.}}&{}&{s}&{2s}\\

    [tex] \begin{align*}K_{sp} &= [Pb^{2+}][Cl^-]^2 \\
    1.2 \times 10^{-5} &= (s)(2s)^2 \\

    Solve for s, that is [Pb2+], the solubility of PbCl2 in distilled water.

    In 0.100 NaCl solution, [Cl-] = 0.100. Do the same as above but this time there is an initial concentration of Cl-. You need to assume that 0.100 + 2s is approximately equal to 0.100. Check the assumption is valid by plugging in the value for s you obtain in the right-hand side of the Ksp equation to see if it equals Ksp.
    Last edited: Dec 19, 2005
  4. Dec 19, 2005 #3
    ok i dont get what i have to do do i do dissociation of NaCl next?

    NaCl--> [tex] Na^+^2 +Cl^- [/tex]

    initial concentration Cl=0.100
    initial concentration Na=0
    change for Cl=x
    change for Na=x
    equilibrium for Na =x
    equilibrium for Cl=(0.100+x)

    where did u get the (0.100+2x) there is no 2 in the equation for dissociation of NaCl.

    what do we do after this? I dont get it where does the distilled water come in
  5. Dec 19, 2005 #4
    NaCl fully dissociates: [Cl-] = [NaCl] = 0.100 mol/L.

    [tex] \begin{array}{ c|ccc|}{}&{PbCl_2 \, \Leftrightarrow}&{Pb^{2+}}&{2Cl^-}\\
    \hline{\text{Initial conc.}}&{}&0&0.100\\
    \text{Change in conc.}}&{}&{}&{}\\
    \text{Conc. at equi.}}&{}&{}&{}\\
    Last edited: Dec 19, 2005
  6. Dec 19, 2005 #5
    Ok but i still dont get how to solve the question
  7. Dec 19, 2005 #6
    So you have
    [tex]1.2 \times 10^{-5} = (s)(0.100 + 2s)^2[/tex]

    We assume that [tex]0.100 + 2s \approx 0.100[/tex] because the Ksp value is quite small and so 's' should have a minimal numerical effect on 0.100.

    So the equation becomes
    [tex]1.2 \times 10^{-5} = (s)(0.100)^2[/tex]

    This assumption must be checked by plugging the value for s (ie. [Pb2+]) into the Ksp equation.
  8. Dec 19, 2005 #7
    so 1.2*10^-5= [0.0144] [0.100]^2?

    1.2*10^-5= 1.44*10^-4
  9. Dec 19, 2005 #8
    Just an arithmetic thing

    [tex]1.2 \times 10^{-5} = s(0.100)^2 = 0.01s[/tex]

    [tex]s = \frac{1.2 \times 10^{-5}}{0.01} = 0.0012[/tex]

    So [tex][Pb^{2+}][Cl^-]^2 = 0.0012 (0.1 + 2\times0.0012) = 1.25 \times 10^{-5} \approx 1.3 \times 10^{-5}[/tex]

    Which isn't good, is it?!

    We have to go back to our unapproximated equation, expand, and solve the cubic equation for s with this site:
    if you don't have access to an alternative.
  10. Dec 20, 2005 #9
    i dont know if we are doing this right its getting complicated i dont think ive done this before
  11. Dec 20, 2005 #10
    The process is fine; the numbers just turned out to bite us. If solving the cubic is too complicated, stick with 0.0012mol/L. The actual value is 0.0011 but this won't make a huge difference when you compare with the solubility in distilled water, which was the goal of the exercise.

    You may be expected to just assume the assumption is valid and move on.
  12. Dec 20, 2005 #11
    i dont know how to compare to the solubility of water. Lol im so confused if you could take me through this step by step it would be appreciated thanks
  13. Dec 20, 2005 #12
    You found the solubility of PbCl2 in distilled water to be 0.0144mol/L (I saw in another thread).

    You found the solubility of PbCl2 in 0.100M NaCl solution to be 0.0012mol/L.

    So the solubility of PbCl2 in distilled water is 0.0144/0.0012 = 12 times greater than the solubility in the NaCl solution.

    This is called the common ion effect.
  14. Dec 20, 2005 #13
    is 0.014 mol/L the correct concentration for the solubility of PbCl2? How do you know that is in distilled water?

    and how did you get 0.0012 mol/L NaCl solution?
  15. Dec 20, 2005 #14
    Because there wasn't a common ion; scroll up to the top of the page.
    The solubility of PbCl2 in the NaCl solution (which itself had a concentration of 0.100mol/L) was 0.0012 mol/L.
  16. Dec 20, 2005 #15


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    adding to what unco has said so far

    Lead chloride would dissolve less in sodium chloride than in distilled water, due to the common ion effect. The complexity of this problem depends on whether it's for general chemistry or analytical chemistry. For now I'll assume general chemistry

    NaCl is assumed to dissociate completely, thus your original concentration of chloride is .100 M, the expression for the dissociation of lead chloride is as follows

    ksp=[x][2x+.100M]^2, solve for x.
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