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Will \(\mathrm{PbCl}_{2}(\mathrm{s})\) precipitate when \(155 \mathrm{mL}\) of \(0.016 \mathrm{M}\) \(\mathrm{KCl}(\mathrm{aq})\) are added to \(245 \mathrm{mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) ?\)

Short Answer

Expert verified
To determine if \(\mathrm{PbCl}_{2}\) will precipitate, calculate the reaction quotient Q from the concentrations of Pb2+ and Cl- ions. If Q > Ksp of PbCl2, then \(\mathrm{PbCl}_{2}\) will precipitate.

Step by step solution

01

- Calculate the new concentration of Cl-

First of all, we need to find the concentration of Cl- ions in the mixed solution. Given that KCl is a strong electrolyte, every one molecule of KCl breaks down into its ions completely, so the Cl- concentration is the same as the KCl concentration. Since the total volume of the solution is (155mL + 245mL) = 400 mL, the new concentration of Cl- ions would be \((155 mL) \times(0.016 M) /(400 mL)\).
02

- Calculate the new concentration of Pb2+

In a similar manner, since Pb(NO3)2 is a strong electrolyte, the concentration of Pb2+ ions is the same as Pb(NO3)2 concentration. So the new concentration of Pb2+ ions would be \((245 mL) \times (0.175 M) /(400 mL)\).
03

- Calculate the reaction quotient (Q)

Now we calculate the reaction quotient (Q) which is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants, raised to their stoichiometric coefficients. Here, since the reaction forming the precipitate is \(\mathrm{Pb^{2+}}(\mathrm{aq})+\mathrm{2Cl^-}(\mathrm{aq}) \rightleftharpoons \mathrm{PbCl}_2(s) \), we have Q = [Pb2+][Cl-]2. Substitute the calculated values for the concentrations of Pb2+ and Cl- ions into this formula to find the value of Q.
04

- Compare Q and Ksp of PbCl2

The solubility product constant (Ksp) of PbCl2 is \(1.7 \times 10^{-5}\). If the calculated Q is more than this Ksp, the PbCl2 will precipitate, as the solution is supersaturated. But if Q is less than or equal to Ksp, PbCl2 will not precipitate, as the solution can accommodate more PbCl2 without reaching saturation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
The Solubility Product Constant, abbreviated as Ksp, is a crucial concept in understanding precipitation reactions. It is an equilibrium constant that applies to the dissolution of a sparingly soluble ionic compound. The Ksp refers to the maximum product of the ion concentrations that can be aloft in solution at equilibrium. For a given sparingly soluble salt, such as lead chloride, PbCl鈧, this constant gives insight into how much of the compound can dissolve before it starts to precipitate.
In the equation for the dissolving of lead chloride:
  • PbCl鈧(s) 鈫 Pb虏鈦(aq) + 2Cl鈦(aq)
Here, the Ksp expression is set up based on the concentrations of the ions at equilibrium:
  • Ksp = [Pb虏鈦篯[Cl鈦籡虏
Each concentration in the expression is raised to the power of its coefficient in the balanced equation. The solubility product helps predict whether a solid will form in a reaction. By comparing the value of the Reaction Quotient (Q) with the Ksp, we can deduce if the solution is saturated (equilibrium), unsaturated (more solid can dissolve), or supersaturated (precipitation will occur). Thus, in our exercise example, if Q exceeds the Ksp of 1.7 x 10鈦烩伒 for PbCl鈧, precipitation takes place.
Electrolyte Solutions
In this context, the concept of Electrolyte Solutions plays an important role. Electrolytes are substances that dissolve in water to form ions, and the solution thus produced is capable of conducting electricity. Electrolytes can be strong or weak, depending on their ability to dissociate into ions.
In the exercise at hand, both KCl and Pb(NO鈧)鈧 dissolve completely in water as strong electrolytes:
  • KCl dissociates completely into K鈦 and Cl鈦 ions.
  • Pb(NO鈧)鈧 dissociates into Pb虏鈦 and NO鈧冣伝 ions.
Strong electrolytes like these ensure that the concentration of the ions in the solution matches the initial concentration of the substance dissolved. This complete dissociation is key in calculating ion concentrations after mixing different solutions. When KCl and Pb(NO鈧)鈧 are mixed, we need to calculate the new concentrations to determine the potential for precipitation.
Understanding how strong electrolytes work is fundamental to working out these calculations and predicting whether reactions such as precipitations might happen.
Reaction Quotient (Q)
To grasp the potential of a precipitation reaction occurring, the Reaction Quotient (Q) is another essential tool. Q is a value obtained much like Ksp, but it is calculated from the current concentrations of the ions in a solution, not necessarily at equilibrium.
In the task of determining whether PbCl鈧 will precipitate or not:
  • First, calculate the concentrations of the individual ions once the solutions are mixed.
  • Then, use those concentrations to find Q using the formula: \[Q = [Pb^{2+}][Cl^{-}]^{2}\]
Once you've computed Q, compare it with the already known Ksp of PbCl鈧:
  • If Q > Ksp, the system is supersaturated, and the excess ions will precipitate as PbCl鈧.
  • If Q = Ksp, the solution is at equilibrium, and no more PbCl鈧 will dissolve.
  • If Q < Ksp, the solution is unsaturated, meaning more PbCl鈧 can still dissolve without forming a precipitate.
Hence, the comparison between Q and Ksp helps chemists predict and explain whether or not a precipitate will form in given reaction circumstances.

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Most popular questions from this chapter

How would you expect the presence of each of the following solutes to affect the molar solubility of \(\mathrm{CaCO}_{3}\) in water: (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3} ;\) (b) \(\mathrm{HCl} ;\) (c) \(\mathrm{NaHSO}_{4}\) ? Explain.

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Which of the following would be most effective, and which would be least effective, in reducing the concentration of the complex ion \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) in a solution: \(\mathrm{HCl}, \mathrm{NH}_{3},\) or \(\mathrm{NH}_{4} \mathrm{Cl} ?\) Explain your choices.

The solubility of \(\mathrm{CdCO}_{3}(\mathrm{s})\) in \(1.00 \mathrm{M} \mathrm{KI}(\mathrm{aq})\) is \(1.2 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Given that \(K_{\mathrm{sp}}\) of \(\mathrm{CdCO}_{3}\) is \(5.2 \times 10^{-12},\) what is \(K_{\mathrm{f}}\) for \(\left[\mathrm{CdI}_{4}\right]^{2-} ?\)

The best way to ensure complete precipitation from saturated \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) of a metal ion, \(\mathrm{M}^{2+}\), as its sulfide, \(\mathrm{MS}(\mathrm{s}),\) is to \((\mathrm{a})\) add an acid; \((\mathrm{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{S}\right]\) in the solution; (c) raise the \(\mathrm{pH} ;\) (d) heat the solution.

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