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Precipitation of lead hydroxide in the presence of EDTA
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[QUOTE="TheSodesa, post: 5455451, member: 512094"] Ok, so \begin{align*} [Pb^{2+}] &= \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} \end{align*} 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. [/QUOTE]
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Precipitation of lead hydroxide in the presence of EDTA
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