/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A 1.00-L solution saturated at \... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 52

A 1.00-L solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2}\). Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility product constant (Ksp) of lead(II) iodide (\(PbI_2\)) at \(25^{\circ} \mathrm{C}\) can be calculated by first finding its molar solubility (0.00117 mol/L), then using the balanced chemical equation to determine the concentrations of the ions in the solution ([Pb虏鈦篯 = 0.00117 mol/L, [I鈦籡 = 0.00234 mol/L), and finally applying the formula Ksp = [Pb虏鈦篯[I鈦籡虏. The Ksp for \(PbI_2\) at \(25^{\circ} \mathrm{C}\) is 6.44 x 10鈦烩伓.

Step by step solution

01

Calculate the molar solubility of PbI鈧�

To find the molar solubility of PbI鈧�, we first need to convert the 0.54g of PbI鈧� to moles. We can do this using its molar mass, which can be found by adding 207.2 g/mol (for lead) and 253.8 g/mol (for two iodine atoms). Molar mass of PbI鈧� = 207.2 g/mol + 253.8 g/mol = 461.0 g/mol Now, we can find the number of moles of PbI鈧�: Moles of PbI鈧� = (0.54 g) / (461.0 g/mol) = 0.00117 mol Since the volume of the solution is 1.00 L, the molar solubility (S) will be equal to the number of moles of PbI鈧�: S = 0.00117 mol/L
02

Write the balanced chemical equation for the dissolution of PbI鈧�

The balanced chemical equation for the dissolution of PbI鈧� is: \(PbI_{2} \rightleftharpoons Pb^{2+} + 2I^-\) When 1 mole of lead(II) iodide dissolves in water, it produces 1 mole of lead(II) ions (Pb虏鈦�) and 2 moles of iodide ions (I鈦�).
03

Calculate the concentrations of Pb虏鈦� and I鈦� ions

From the balanced chemical equation, we can see that: 1 mole of PbI鈧� 鈫� 1 mole of Pb虏鈦� and 2 moles of I鈦� 0.00117 mol/L of PbI鈧� 鈫� 0.00117 mol/L of Pb虏鈦� and 0.00234 mol/L of I鈦� Now we have the concentrations of both ions in the solution: [Pb虏鈦篯 = 0.00117 mol/L [I鈦籡 = 0.00234 mol/L
04

Calculate the solubility product constant (Ksp)

Now that we have the concentrations of Pb虏鈦� and I鈦� ions, we can find the solubility product constant (Ksp) using the formula: Ksp = [Pb虏鈦篯[I鈦籡虏 Ksp = (0.00117)(0.00234)虏 = 6.44 x 10鈦烩伓 So, the solubility product constant for lead(II) iodide (PbI鈧�) at 25掳C is 6.44 x 10鈦烩伓.

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Most popular questions from this chapter

Use information from Appendix D to calculate the \(\mathrm{pH}\) of (a) a solution that is \(0.060 \mathrm{M}\) in potassium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOK}\right.\) or \(\left.\mathrm{KC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) and \(0.085 \mathrm{Min}\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right) ;\) (b) a solution that is \(0.075 \mathrm{M}\) in trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\), and \(0.10 \mathrm{M}\) in trimethylammonium chloride, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NHCl} ;\) (c) a solution that is made by mixing \(50.0 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) acetic acid and \(50.0 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) sodium acetate.

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C}\) ?

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH}\), (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3}\) (c) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), (e) \(\mathrm{CuBr}\).

(a) Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (HCl) is mixed with a solution of sodium formate \(\left(\mathrm{NaCHO}_{2}\right)\). (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equilibrium concentrations of \(\mathrm{Na}^{+}, \mathrm{Cl}^{-}\), \(\mathrm{H}^{+}, \mathrm{CHO}_{2}^{-}\), and \(\mathrm{HCHO}_{2}\) when \(50.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{HCl}\) is mixed with \(50.0 \mathrm{~m}\) L of \(0.15 \mathrm{M} \mathrm{NaCHO}_{2}\).
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