QuestionJuly 15, 2025

Pb_(3)PO_(4) has a molar solublity of 3.0^ast 10^-23 What is the molar solubility in a solution of 0.0230MNa_(3)PO_(4) Answer

Pb_(3)PO_(4) has a molar solublity of 3.0^ast 10^-23 What is the molar solubility in a solution of 0.0230MNa_(3)PO_(4) Answer
Pb_(3)PO_(4) has a molar solublity of 3.0^ast 10^-23 What is the molar solubility in a solution of 0.0230MNa_(3)PO_(4) Answer

Solution
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Answer

s \approx 1.6 \times 10^{-9} M Explanation 1. Write the solubility product expression For Pb_3(PO_4)_2, the dissociation is Pb_3(PO_4)_2 \rightarrow 3Pb^{2+} + 2PO_4^{3-}. The solubility product K_{sp} is given by K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2. 2. Determine initial concentrations In a solution of 0.0230M Na_3PO_4, the concentration of PO_4^{3-} ions is 0.0230M. 3. Set up the equilibrium expression Let s be the molar solubility of Pb_3(PO_4)_2 in the presence of 0.0230M PO_4^{3-}. Then, [Pb^{2+}] = 3s and [PO_4^{3-}] = 0.0230 + 2s. Substitute these into the K_{sp} expression: K_{sp} = (3s)^3(0.0230 + 2s)^2. 4. Simplify and solve for s Given K_{sp} = 3.0 \times 10^{-23}, assume 2s is negligible compared to 0.0230: K_{sp} \approx (3s)^3(0.0230)^2. Solve for s: s = \left(\frac{K_{sp}}{27 \times 0.0230^2}\right)^{1/3}. 5. Calculate the molar solubility Plug in the values: s = \left(\frac{3.0 \times 10^{-23}}{27 \times 0.0230^2}\right)^{1/3}.

Explanation

1. Write the solubility product expression<br /> For $Pb_3(PO_4)_2$, the dissociation is $Pb_3(PO_4)_2 \rightarrow 3Pb^{2+} + 2PO_4^{3-}$. The solubility product $K_{sp}$ is given by $K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2$.<br /><br />2. Determine initial concentrations<br /> In a solution of $0.0230M$ $Na_3PO_4$, the concentration of $PO_4^{3-}$ ions is $0.0230M$.<br /><br />3. Set up the equilibrium expression<br /> Let $s$ be the molar solubility of $Pb_3(PO_4)_2$ in the presence of $0.0230M$ $PO_4^{3-}$. Then, $[Pb^{2+}] = 3s$ and $[PO_4^{3-}] = 0.0230 + 2s$. Substitute these into the $K_{sp}$ expression: $K_{sp} = (3s)^3(0.0230 + 2s)^2$.<br /><br />4. Simplify and solve for $s$<br /> Given $K_{sp} = 3.0 \times 10^{-23}$, assume $2s$ is negligible compared to $0.0230$: $K_{sp} \approx (3s)^3(0.0230)^2$. Solve for $s$: $s = \left(\frac{K_{sp}}{27 \times 0.0230^2}\right)^{1/3}$.<br /><br />5. Calculate the molar solubility<br /> Plug in the values: $s = \left(\frac{3.0 \times 10^{-23}}{27 \times 0.0230^2}\right)^{1/3}$.
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