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Biology and Chemistry Homework Help
Precipitation of lead hydroxide in the presence of EDTA
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[QUOTE="TheSodesa, post: 5454807, member: 512094"] [h2]Homework Statement [/h2] [B]1. The question[/B] 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? \begin{array}{|l|c|} 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 \end{array} [h2]Homework Equations[/h2] 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. [h2]The Attempt at a Solution[/h2] 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 [I]this[/I] reaction then reacts with the ##OH^-##-ions. Below is a table of the first reaction. \begin{array}{|l|c|} concentration & Pb^{2+} & EDTA^{4-} & PbEDTA^{2-}\\ \hline beginning & 0.01 & 0.05 & 0\\ end & 0.01-x & 0.05-x & x \end{array} We can then solve for ##x## using ##K##: \begin{align*} K &= \frac{x}{(0.01-x)(0.05-x)}\\ \iff\\ &(0.01-x)(0.05-x)K = x\\ \iff\\ &\frac{1}{2000}K-0.06Kx + Kx^2 = x\\ \iff\\ & Kx^2 -(0.06K+1)x + \frac{1}{2000}K = 0 \end{align*} Solving for ##x##: \begin{align*} x &= \frac{(0.06K+1) \pm \sqrt{(0.06K+1)^2 - 4K\times \frac{K}{2000}})}{2K}\\ &= \begin{cases} 1/20\\ 1/100 \end{cases} \end{align*} 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. [/QUOTE]
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Precipitation of lead hydroxide in the presence of EDTA
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