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Homework Help: Precipitation of lead hydroxide in the presence of EDTA

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data

    1. The question

    EthyleneDiamineTetra-acetate(##EDTA^{4-}##) is used in chemical analysis as a complex-former. ##EDTA^{4-}## is also used to treat heavy metal poisoning, as it forms water soluble complexes with them which can then be easily removed by the body. The reaction of ##EDTA^{4-}## with lead is:
    $$Pb^{2+}(aq) + EDTA^{4-}(aq) \rightleftharpoons PbEDTA^{2-}(aq), K = 1.1 \times 10^{18}$$

    0.010 mol of ##Pb(NO_3)_2##-solution is added to 1 liter ##(dm^3)## of buffered water solution (pH = 13), whose ##Na_4 EDTA## concentration is 0.050M. Does the added lead precipitate in the form of ##Pb(OH)_2##-salt?

    Some \ values\\ \hline
    K_s(Pb(OH)_2) & 1.2 \times 10^{-15}\\
    K & 1.1\times 10^{18}\\
    n_0 (Pb(NO_3)_2) & 0.010 mol\\
    [Na_4 (EDTA)]_0 & 0.050M\\
    pH & 13

    2. Relevant equations

    If our chemical reaction is
    $$aA + bB \rightleftharpoons cC + dD $$
    then our equilibrium constant
    $$K = \frac{[A]^a [ B ]^b}{[C]^c[D]^d} $$
    The solubility constant would be
    $$K_s = [ C ]^c [D]^d$$
    since the activity ##a_s## of the pure solid on the left side of the reaction equation is 1.

    3. The attempt at a solution

    I managed to solve this, I just need a second opinion on whether my solution is correct.

    The lead is going to precipitate if the reaction quotient
    $$Q > K_s$$
    To see if this holds, I need to find out the concentration of lead in the solution. I'm going to assume that the added ##Pb(NO_3)_2## dissolves completely and then reacts with the ##EDTA^{4-}##, before the remaining lead from this reaction then reacts with the ##OH^-##-ions. Below is a table of the first reaction.
    concentration & Pb^{2+} & EDTA^{4-} & PbEDTA^{2-}\\ \hline
    beginning & 0.01 & 0.05 & 0\\
    end & 0.01-x & 0.05-x & x
    We can then solve for ##x## using ##K##:
    &= \frac{x}{(0.01-x)(0.05-x)}\\
    &(0.01-x)(0.05-x)K = x\\
    &\frac{1}{2000}K-0.06Kx + Kx^2 = x\\
    & Kx^2 -(0.06K+1)x + \frac{1}{2000}K = 0
    Solving for ##x##:
    &= \frac{(0.06K+1) \pm \sqrt{(0.06K+1)^2 - 4K\times \frac{K}{2000}})}{2K}\\
    Now ##1/20 = 0.05## won't do, since ##(0.01-x)## would be negative. Plugging in ##x = 0.01## gives us the concentration of lead after the reaction which is ##[Pb]_1 = (0.01 - 0.01) = 0##. Therefore at these concentrations there is no lead left over to react with the ##OH^-##-ions, and there can be no precipitate.

    This is a desirable result, considering the use of EDTA in medicine, but this seems too easy. I must be missing something or making wrong assumptions.
    Last edited: Apr 25, 2016
  2. jcsd
  3. Apr 26, 2016 #2


    User Avatar

    Staff: Mentor

    No, concentration of Pb2+ is not zero. Both 1/20 and 1/100 are only approximations. What they do tell you though is that the Pb2+ is complexed almost entirely, with concentration of the complexed form equal - for all practical purposes - 0.01 M. Plug this number (together with the concentration of the free EDTA4- left after the reaction) into the K equation and calculate equilibrium concentration of the Pb2+.
  4. Apr 26, 2016 #3
    If I understood you correctly, I'm supposed to invoke Hess's law, meaning one can assume all of the lead reacts with EDTA and then the reverse reaction happens, but then instead of using
    $$K = \frac{[PbEDTA^{2-}]}{[Pb^{2+}][EDTA^{4-}]},$$
    I should use the equilibrium constant of the opposite reaction
    $$K_o = \frac{[Pb^{2+}][EDTA^{4-}]}{[PbEDTA^{2-}]} = \frac{1}{K}.$$
    Then if ##y## is the concentration of lead after the reaction,
    concentration & [PbEDTA^{2-}] & [Pb^{2+}] & [EDTA^{4-}] \\ \hline
    beginning & 0.01 & 0 & 0.04\\
    end & 0.01 - y & y & 0.04 + y\\
    &= \frac{1}{K} = \frac{(y) (0.04 + y)}{0.01 - y}\\
    & 0.01 - y = (y)(0.04 + y)K\\
    & 0.01 - y = 0.04Ky + Ky^2\\
    & Ky^2 + (0.04K + 1)y - 0.01 = 0
    Solving for y
    &= \frac{-(0.04K + 1) \pm \sqrt{(0.04K + 1)^2 - 4K \times 0.01}}{2K}\\
    Again, ##\frac{-1}{25}## obviously won't do, so the concentration of lead after the reverse reaction is 0. Unless I plugged something in incorrectly, I still don't get any lead.

    Ninja edit: I let Wolfram Alpha do the math, and instead of doing what my cheap-o pocket calculator did (rounding to zero), it produced ##y = 2.27273 \times 10^{-19}##.

    This allows us to calculate the value of the reaction quotient $$Q = [Pb^{2+}] [OH^{-}]^2$$
    Here $$[OH^{-}] = 10^{-1},$$
    $$pH + pOH = 14 \iff pOH = 14 - pH = 1$$
    $$pOH = -lg[OH^{-}]$$
    &= [Pb^{2+}] [OH^{-}]^2\\
    &= (2.27273 \times 10^{-19})(10^{-1})^2\\
    &= 2.27273 \times 10^{-21},\\
    meaning ##Q<<K_s## and the lead won't precipitate.

    EDIT2: I got my ##K## wrong in the OP, it seems (reversed order of products and reactants). I blame trying to type in ##LaTeX##.
    Last edited: Apr 26, 2016
  5. Apr 26, 2016 #4


    User Avatar

    Staff: Mentor

    No idea why you invoke the Hess's law, it is completely unrelated.

    No need for ICE table here, nor most of the things you did. You have correctly stated that

    [tex]K = \frac{[PbEDTA^{2-}]}{[Pb^{2+}][EDTA^{4-}]}[/tex]

    Just solve for [Pb2+] and plug in concentrations of PbEDTA2- as 0.01 M and concentration of EDTA4- as 0.04 M. This is not exact, but good enough. 0.01-2.3×10-19 is close enough to 0.01 that the exact number doesn't matter. Tr to do teh calculations the way I told you and you will see for yourself.

    This kind of approximation is often a must when dealing with equilibrium calculations, as a rule of thumb if you add or subtract two numbers and one is smaller than 5% of the other, you can ignore it and see if the results make sense. That is 1+0.05≈1-0.05≈1.
  6. Apr 26, 2016 #5
    Ok, so
    &= \frac{[PbEDTA^{2-}]}{K[EDTA^{4-}]}\\
    &= \frac{0.01}{(1.1 \times 10^{18})(0.04)}\\
    &= 2.\overline{27} \times 10^{-19}\\
    &\approx 2.27 \times 10^{-19}
    Yeah, the exact same result.

    As for Hess's law, I was under the impression, that the reason we can make the assumption that the first reaction goes to completion and then returns to the wanted state while still producing the desired value is because enthalpy is a state function.
  7. Apr 26, 2016 #6


    User Avatar

    Staff: Mentor

    Enthalpy is not enough to describe the reaction.
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