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Chapter 16: Problem 46

Calculate the solubility of solid $\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1 \times 10^{-54}\right)\( in a \)0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ solution.

Short Answer

Expert verified
The solubility of Pb3(PO4)2 in a 0.10 M solution of Pb(NO3)2 is \(1.0\times10^{-17}\) M.

Step by step solution

01

Write the solubility product expression

The solubility product expression for Pb3(PO4)2 is given by: \[K_{sp} = [Pb^{2+}]^3 [PO_{4}^{3-}]^2\]
02

Setup the ICE table

| | Pb^{2+} | PO_{4}^{3-} | |---------|---------|-----------| |Initial | 0.10 | 0 | |Change | + 3x | + 2x | |Equilibrium| 0.10+3x| 2x |
03

Substitute the equilibrium concentrations into the solubility product expression

Using the equilibrium concentrations from the ICE table, substitute them into the solubility product expression: \[K_{sp} = (0.10+3x)^3 (2x)^2\] Since Ksp is given as \(1\times10^{-54}\), we have: \(1\times10^{-54}=(0.10+3x)^3 (2x)^2\)
04

Solve for x

To solve for x, we will first simplify the equation and then solve for x using a numerical method or approximation. Due to the very low value of Ksp, it is reasonable to assume that \(3x<< 0.10\) and that the 3x term in the equation can be ignored, which simplifies the equation to: \(1\times10^{-54}=(0.10)^3 (2x)^2\) Now we can solve for x: \(x = \sqrt{\frac{1\times10^{-54}}{0.10^3 \times 4}}\) \(x = 5.0\times10^{-18}\)
05

Calculate the solubility of Pb3(PO4)2

Now that we have found the value of x, we can use it to find the solubility of Pb3(PO4)2. Since 2x moles of PB3(PO4)2 dissolve for each mole of PO4^{3-}, the molar solubility of Pb3(PO4)2 is given by 2x: Solubility of Pb3(PO4)2 \(= 2 \times 5.0\times10^{-18} = 1.0\times10^{-17}\) M Therefore, the solubility of Pb3(PO4)2 in a 0.10 M solution of Pb(NO3)2 is \(1.0\times10^{-17}\) M.

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Most popular questions from this chapter

Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.

Calculate the molar solubility of $\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}.$

Calculate the solubility of solid $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\( in a \)0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}$ solution.

Silver cyanide \((\mathrm{AgCN})\) is an insoluble salt with \(K_{\mathrm{sp}}=2.2 \times 10^{-12}\) . Compare the effects on the solubility of silver cyanide by addition of \(\mathrm{HNO}_{3}(a q)\) or by addition of \(\mathrm{NH}_{3}(a q).\)

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of “nano machines.” One common building block of nano structured architectures is manganese oxide nano particles. The particles can be formed from manganese oxalate nano rods, the formation of which can be described as follows: $$\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}, \mathrm{O}_{4}(a q) \quad K_{1}=7.9 \times 10^{3}$$ $$\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q) \quad K_{2}=7.9 \times 10^{1}$$ Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-} :\) $$K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}$$
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