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Magazine Chapter 16: Problem 56
A solution is prepared by mixing \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{~mL}\) of \(1.0 M \mathrm{KCl}\). Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\mathrm{sp}}\right.\) for \(\mathrm{PbCl}_{2}(s)\) is \(\left.1.6 \times 10^{-5} .\right]\)
Short Answer
Expert verified
The equilibrium concentrations in the solution are [Pb虏鈦篯 = \(1.0 \times 10^{-5}\:M\) and [Cl鈦籡 = 0.4 M.
Step by step solution
01
Write the balanced chemical equation for the reaction
First, we need to write the balanced chemical equation for the reaction when the two solutions are mixed. The reaction involves the formation of solid lead(II) chloride (PbCl鈧�) from aqueous lead(II) nitrate (Pb(NO鈧�)鈧�) and aqueous potassium chloride (KCl):
Pb(NO鈧�)鈧�(aq) + 2KCl(aq) 鈫� PbCl鈧�(s) + 2KNO鈧�(aq)
02
Calculate the initial concentrations in moles
Next, we need to find the initial moles of both reactants in the mixture. We will first determine the moles of Pb虏鈦� ions and Cl鈦� ions.
For Pb虏鈦� ions:
Moles = Volume 脳 Molarity
Moles = \(50.0\:mL 脳 0.10\:M\)
Moles = \(5.0 \times 10^{-3}\:mol\)
For Cl鈦� ions:
Moles = Volume 脳 Molarity
Moles = \(50.0\:mL 脳 1.0\:M\)
Moles = \(50.0 \times 10^{-3}\:mol\)
03
Determine the reaction stoichiometry and find the equilibrium concentrations
Now, we need to determine how much Pb虏鈦� and Cl鈦� ions reacted according to stoichiometry. According to the balanced equation, 1 mole of Pb虏鈦� ions reacts with 2 moles of Cl鈦� ions. Therefore, both moles will not be consumed completely.
From the stoichiometry of the reaction, we can calculate the moles of Cl鈦� ions that reacted with Pb虏鈦� ions:
Moles of Cl鈦� ions used = 2 脳 moles of Pb虏鈦� ions
Moles of Cl鈦� ions used = 2 脳 \(5.0 \times 10^{-3}\:mol\)
Moles of Cl鈦� ions used = \(10.0 \times 10^{-3}\:mol\)
After the reaction occurs, we will have the following moles:
Moles of Cl鈦� ions remaining = initial moles of Cl鈦� ions - moles of Cl鈦� ions used
Moles of Cl鈦� ions remaining = \((50.0 - 10.0) \times 10^{-3}\:mol\)
Moles of Cl鈦� ions remaining = \(40.0 \times 10^{-3}\:mol\)
Now, we have no moles of Pb虏鈦� ions remaining in solution since it has all precipitated as PbCl2.
Next, we need to find the equilibrium concentrations using the total volume of the solution, which is 100 mL:
\[Equilibrium\_Concentration\ of [Cl^-] = \frac{Moles\ of\ Cl^- \:ions}{Volume\ of\ solution}\]
\[Equilibrium\_Concentration\ of [Cl^-] = \frac{40.0 \times 10^{-3}\:mol}{100.0\:mL}\]
\[Equilibrium\_Concentration\ of [Cl^-] = 0.4\:M\]
The concentration of Pb虏鈦� ions in the solution can be found using Ksp:
\[K_{sp} = [Pb^{2+}] [Cl^-]^2\]
\[1.6 \times 10^{-5} = [Pb^{2+}] (0.4)^2\]
Now, solve for [Pb虏鈦篯:
\[ [Pb^{2+}] = \frac{1.6 \times 10^{-5}}{(0.4)^2}\]
\[ [Pb^{2+}] = 1.0 \times 10^{-5}\]
04
Present the final result
Now, we have the equilibrium concentrations of Pb虏鈦� and Cl鈦� ions in the solution:
[Cl鈦籡 = 0.4 M
[Pb虏鈦篯 = \(1.0 \times 10^{-5}\:M\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant
The solubility product constant (\( K_{sp} \) ) is a vital concept in understanding chemical equilibrium in terms of solubility. It helps in determining whether a certain compound will precipitate in a solution.
When a slightly soluble ionic compound dissolves in water, it usually dissociates into its respective ions. The solubility product is the product of the molar concentrations of these ions, each raised to the power of its coefficient in the balanced chemical equation.
In the case of lead(II) chloride (\( PbCl_2 \)), which dissociates as \( PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \), the \( K_{sp} \) can be represented as:
When a slightly soluble ionic compound dissolves in water, it usually dissociates into its respective ions. The solubility product is the product of the molar concentrations of these ions, each raised to the power of its coefficient in the balanced chemical equation.
In the case of lead(II) chloride (\( PbCl_2 \)), which dissociates as \( PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \), the \( K_{sp} \) can be represented as:
- \( K_{sp} = [Pb^{2+}][Cl^-]^2 \)
Reaction Stoichiometry
Stoichiometry in reactions involves using a balanced equation to calculate the proportions of reactants and products. It's essential for determining how much of each substance is needed or produced during a reaction.
For the reaction where lead(II) nitrate reacts with potassium chloride to form lead(II) chloride and potassium nitrate:
Stoichiometry helps us determine the exact amounts of product formed and reactants consumed. By understanding how the coefficients relate, we can calculate remaining concentrations after the reaction, considering any limiting reactants.
For the reaction where lead(II) nitrate reacts with potassium chloride to form lead(II) chloride and potassium nitrate:
- \( Pb(NO_3)_2(aq) + 2KCl(aq) \rightarrow PbCl_2(s) + 2KNO_3(aq) \)
Stoichiometry helps us determine the exact amounts of product formed and reactants consumed. By understanding how the coefficients relate, we can calculate remaining concentrations after the reaction, considering any limiting reactants.
Molarity
Molarity is a fundamental way of expressing concentration in chemistry, defined as the number of moles of a solute per liter of solution. It allows us to determine how concentrated a solution is and is represented by the unit \( M \).
In this problem, molarity is used to calculate the initial concentrations of solutions before they are mixed:
In this problem, molarity is used to calculate the initial concentrations of solutions before they are mixed:
- For lead(II) nitrate: \( 0.10 \, M \)
- For potassium chloride: \( 1.0 \, M \)
Precipitation Reactions
Precipitation reactions involve the formation of a solid from two aqueous solutions. They are evidence of chemical reactions occurring when ions in solution form an insoluble compound.
In our example, mixing lead(II) nitrate and potassium chloride results in the formation of lead(II) chloride (\( PbCl_2 \)) as a precipitate. A solid forms when the product of ionic concentrations exceeds the solubility product constant.
These reactions are useful in analytical chemistry for isolating and purifying substances since the precipitate can be filtered out. Understanding the conditions under which precipitation occurs helps chemists manipulate reactions and predict the solubility of compounds in various solutions. Studying the stoichiometry and using \( K_{sp} \) helps us decide when and how effectively a precipitate will form in specific conditions.
In our example, mixing lead(II) nitrate and potassium chloride results in the formation of lead(II) chloride (\( PbCl_2 \)) as a precipitate. A solid forms when the product of ionic concentrations exceeds the solubility product constant.
These reactions are useful in analytical chemistry for isolating and purifying substances since the precipitate can be filtered out. Understanding the conditions under which precipitation occurs helps chemists manipulate reactions and predict the solubility of compounds in various solutions. Studying the stoichiometry and using \( K_{sp} \) helps us decide when and how effectively a precipitate will form in specific conditions.
