Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of $\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}$ b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

Step by step solution

01

Write the dissociation equation of \(\mathrm{Pb}(\mathrm{OH})_{2}\)

When \(\mathrm{Pb}(\mathrm{OH})_{2}\) dissociates in water, it forms a \(\mathrm{Pb}^{2+}\) ion and two \(\mathrm{OH}^-\) ions. The dissociation equation for this is: \[\mathrm{Pb}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{OH}^-(aq)\]

02

Calculate the concentration of \(\mathrm{Pb}^{2+}\) and \(\mathrm{OH}^-\)

If 's' is the solubility of \(\mathrm{Pb}(\mathrm{OH})_{2}\) in moles/L, then the concentration of \(\mathrm{Pb}^{2+}\) is 's' and the concentration of \(\mathrm{OH}^-\) is '2s'. We are given \(K_{sp}=1.2\times10^{-15}\), so we can write the equation: \[K_{sp}= [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2 = s(2s)^2 = 4s^3 \]

03

Find the solubility and concentration of \(\mathrm{Pb}^{2+}\)

Now, solve for s: \[s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{1.2\times 10^{-15}}{4}}\] Evaluating the terms gives us s: \[s \approx 1.57 \times 10^{-6} \,\text{mol/L}\] So the concentration of \(\mathrm{Pb}^{2+}\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) is approximately \(1.57 \times 10^{-6}\,\text{M}\). Case B: Saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13\)

04

Calculate the concentration of \(\mathrm{OH}^-\) from the given pH

The relationship between pH and the concentration of \(\mathrm{OH^-}\) ions in a solution is: \(pOH = 14 - pH\) We are given that pH = 13, so pOH is: \(pOH = 14 - 13 = 1\) Now, we can find the concentration of \(\mathrm{OH^-}\): \[[\mathrm{OH}^-] = 10^{-pOH} = 10^{-1} = 0.1\,\text{M}\]

05

Find the concentration of \(\mathrm{Pb}^{2+}\) using \(K_{sp}\)

Now that we have the concentration of \(\mathrm{OH^-}\), we can calculate the concentration of \(\mathrm{Pb}^{2+}\) using \(K_{sp}=1.2\times 10^{-15}\): \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{OH}^-]^2\] \[[\mathrm{Pb}^{2+}] = \frac{K_{sp}}{[\mathrm{OH}^-]^2} = \frac{1.2\times 10^{-15}}{(0.1)^2} = 1.2\times 10^{-13}\,\text{M}\] Case C: Buffered solution with \(\mathrm{Pb}^{2+}\) and \(\mathrm{EDTA}^{4-}\)

06

Write the reaction between \(\mathrm{Pb}^{2+}\) and \(\mathrm{EDTA}^{4-}\)

The reaction between \(\mathrm{Pb}^{2+}\) and \(\mathrm{EDTA}^{4-}\) is given by: \[\mathrm{Pb}^{2+}(aq) + \mathrm{EDTA}^{4-}(aq) \rightleftharpoons \mathrm{PbEDTA}^{2-}(aq)\] with the formation constant, \(K = 1.1\times 10^{18}\).

07

Determine if \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitates in the solution

We are given the concentration of \(\mathrm{Pb}^{2+}\) ions and \(\mathrm{EDTA}^{4-}\) ions in the solution. To determine if \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitates in the solution, we must check if the ion product \((Q)\) in the solution is greater than, equal to or less than the solubility product \((K_{sp})\). If \(Q>K_{sp}\), precipitation occurs, if \(Q=K_{sp}\), the solution is saturated and if \(Q K_{sp} = 1.2 \times 10^{-15}\] Since \(Q > K_{sp}\), \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate in the solution. In summary, the concentration of \(\mathrm{Pb}^{2+}\) is: a) \(1.57 \times 10^{-6}\,\text{M}\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\), b) \(1.2 \times 10^{-13}\,\text{M}\) in a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13\), and c) \(\mathrm{Pb}(\mathrm{OH})_{2}\) will precipitate in the given solution with \(\mathrm{EDTA}^{4-}\).

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