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Chapter 14: Problem 115

The solubility of \(\mathrm{PbI}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is \(1.52 \times 10^{-3}\) M. How many grams of \(\mathrm{PbI}_{2}\) will dissolve in \(2.50 \times 10^{6}\) gallons of water at \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Therefore, \(6.62 \times 10^{6}\) grams of PbI₂ will dissolve in \(2.50 \times 10^{6}\) gallons of water at 25°C.

Step by step solution

01

Convert gallons to liters

We are given the volume of water as 2.50 x 10⁶ gallons. To convert this volume to liters, we will use the following conversion factor: 1 gallon = 3.78541 liters \(Volume\,in\,liters = (2.50 \times 10^{6})\,gallons \times \frac{3.78541\,liters}{1\,gallon} = 9.46 \times 10^{6}\,liters\)
02

Calculate moles of PbI₂ that can dissolve

We know the solubility of PbI₂ in water is 1.52 x 10⁻³ M. Thus, in one liter of water, we have 1.52 x 10⁻³ moles of PbI₂. To find the number of moles in the given volume, we can use the following calculation: Moles of PbI₂ = (solubility in moles per liter) x (volume in liters) Moles of PbI₂ = (1.52 x 10⁻³) M x (9.46 x 10⁶) L ≈ 14402.32 moles
03

Convert moles of PbI₂ to grams

To convert the calculated moles of PbI₂ to grams, we need to multiply the number of moles by the molar mass of PbI₂: Molar mass of PbI₂ = 207.2 g/mol (Pb) + 2 x 126.9 g/mol (I) ≈ 460 g/mol Mass of PbI₂ = moles of PbI₂ x molar mass of PbI₂ Mass of PbI₂ = 14402.32 moles * 460 g/mol ≈ 6.62 x 10⁶ grams Therefore, 6.62 x 10⁶ grams of PbI₂ will dissolve in 2.50 x 10⁶ gallons of water at 25°C.

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Most popular questions from this chapter

Which of the following reactions is described by the equilibrium constant expression \(K_{\mathrm{eq}}=\frac{[\mathrm{A}]^{2} \times[\mathrm{B}]^{3}}{[\mathrm{C}]^{3} \times[\mathrm{D}]^{2}}\) (a) \(\mathrm{A}_{2}+\mathrm{B}_{3} \rightleftarrows \mathrm{C}_{3}+\mathrm{D}_{2}\) (b) \(2 \mathrm{~A}+3 \mathrm{~B} \rightleftarrows 3 \mathrm{C}+2 \mathrm{D}\) (c) \(3 \mathrm{C}+2 \mathrm{D} \rightleftarrows 2 \mathrm{~A}+3 \mathrm{~B}\) (d) \(A^{2}+B^{3} \rightleftarrows C^{3}+D^{2}\) (e) \(2 \mathrm{C}+3 \mathrm{D} \rightleftarrows 3 \mathrm{~A}+2 \mathrm{~B}\)

How does decreasing the temperature affect the value of \(K_{e q}\) for an exothermic reaction? (a) Increases \(K_{\text {eq }}\) (b) Decreases \(K_{\text {eq }}\) (c) Does not change \(K_{\text {eq }}\)

Sparingly soluble \(\mathrm{PbCl}_{2}\) dissolves in water to yield an equilibrium \(\mathrm{Pb}^{2+}(a q)\) concentration of \(0.039 \mathrm{M}\). (a) Write the balanced equilibrium equation for \(\mathrm{PbCl}_{2}(\) s) dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for \(\mathrm{PbCl}_{2}\). (c) What is the equilibrium concentration of chloride ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{PbCl}_{2}\) (show your calculation).

The solubility of iron(II) hydroxide, \(\mathrm{Fe}(\mathrm{OH})_{2}\), is \(1.43 \times 10^{-3} \mathrm{~g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) (a) Write a balanced equation for the solubility equilibrium. (b) Write the expression for \(K_{\mathrm{sp}}\) and calculate its value. (c) \(\mathrm{A} 60.0 \mathrm{~mL}\) sample of \(3.00 \times 10^{-3} \mathrm{M} \mathrm{FeSO}_{4}\) solution is added to \(60.0 \mathrm{~mL}\) of \(4.00 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) solution. Does a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{2}\) form? (Hint: Use the concentration values for \(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\) and plug them into the \(K_{\mathrm{sp}}\) expression. If the value you get is larger than \(K_{\mathrm{sp}}\) ' precipitation will occur.)

At \(25^{\circ} \mathrm{C}\), the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in water is \(2.86 \times 10^{-9} \mathrm{M}\). What are the equilibrium concentrations of the cation and the anion in a saturated solution?
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