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Chapter 17: Problem 67

(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?\) (b) Will \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate when \(100 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(5.0 \times 10^{-2} \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Short Answer

Expert verified
(a) No, Ca(OH)鈧� will not precipitate from the solution because Q < Ksp (\(5 \times 10^{-14} < 6.5 \times 10^{-6}\)). (b) Yes, Ag鈧係O鈧� will precipitate when the solutions mix because Q > Ksp (\(4.24 \times 10^{-7} > 1.12 \times 10^{-5}\)).

Step by step solution

01

Find the concentration of OH-

The pH value is 8.0, so we can calculate the [OH-] concentration using the relationship: \(pOH = 14 - pH\) The pOH is equal to 6, so we can find the concentration of hydroxide ions (OH鈦�) as follows: \[ [OH鈦籡 = 10^{-pOH} = 10^{-6} = 1 \times 10^{-6} \mathrm{M} \]
02

Calculate Ion Product (Q) and compare with Ksp

Now we will use the concentration of Ca虏鈦� and OH鈦� ions to calculate the ion product (Q) for Ca(OH)鈧�. Since the given solution is 0.050 M CaCl鈧�, we have a [Ca虏鈦篯 of 0.050 M. Then, the ion product (Q) for Ca(OH)鈧� is: \[Q = [Ca虏鈦篯[OH鈦籡^2 = (0.050)(1 \times 10^{-6})^2 \] \[Q = 5 \times 10^{-14} \] The Ksp for Ca(OH)鈧� is approximately \(6.5 \times 10^{-6}\). Since Q < Ksp, the system is not supersaturated, and the solution is not concentrated enough for precipitation to occur. (a) Answer: No, Ca(OH)鈧� will not precipitate from the solution. (b) Determine if \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) will precipitate
03

Calculate final concentrations of ions after mixing

We're given 100 mL of 0.050 M AgNO鈧� and 10 mL of 5.0 x 10鈦宦� M Na鈧係O鈧� When these solutions mix, the total volume becomes 110 mL, so we can calculate the final concentrations of the ions using the dilution formula: \[C_{1}V_{1} = C_{2}V_{2}\] For Ag鈦�: \[(0.050)(100) = C_{Ag}(110)\] \[C_{Ag} = \frac{0.050 \times 100}{110} = 4.55 \times 10^{-2} M\] For SO鈧劼测伝: \[(5.0 \times 10^{-2})(10) = C_{SO4}(110)\] \[C_{SO4} = \frac{5.0 \times 10^{-2} \times 10}{110} = 4.55 \times 10^{-3} M\]
04

Calculate Ion Product (Q) and compare with Ksp

We will now use the final concentrations of Ag鈦� and SO鈧劼测伝 to calculate the ion product(Q) for Ag鈧係O鈧�: \[Q = [Ag鈦篯^2[SO鈧刕{2-}] = (4.55 \times 10^{-2})^2(4.55 \times 10^{-3})\] \[Q \approx 4.24 \times 10^{-7}\] The Ksp for Ag鈧係O鈧� is approximately \(1.12 \times 10^{-5}\). Since Q > Ksp, the solution is supersaturated, and Ag鈧係O鈧� will precipitate. (b) Answer: Yes, Ag鈧係O鈧� will precipitate when the solutions mix.

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Key Concepts

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

Precipitation Reactions
Precipitation reactions occur when two ionic solutions are mixed, resulting in the formation of an insoluble solid or precipitate. This process happens when the concentration of the ionic product in the mixture exceeds the solubility limit of the compound, causing ions to come out of solution and form an aggregated solid. To predict whether a precipitation reaction will take place, chemists determine if the product of the concentrations of the ions in the solution, known as the ion product (Q), exceeds the solubility product constant (Ksp). This method is often used to understand various important processes such as water treatment and qualitative chemical analysis.
The steps are straightforward: Assess the concentrations of potential precipitants. Calculate the ion product. Compare the ion product to the Ksp value.
  • If Q > Ksp, precipitation is likely to occur.
  • If Q < Ksp, the mixture remains a homogeneous solution.
  • If Q = Ksp, the solution is in equilibrium, with no net precipitation or dissolving.
Understanding these reactions is crucial for grasping how solubility equilibria in ionic solutions govern the formation of precipitates.
Ion Product (Q)
The ion product (Q) is a vital tool in predicting and understanding solubility equilibria in mixtures. Q represents the current product of the concentrations of the ions in a solution and offers a snapshot of the system's status relative to the equilibrium state. To calculate Q, simply multiply the concentrations of the respective ionic species involved in the potential compound's formation.
This comparison of Q to the Ksp can predict potential precipitation:
  • When Q < Ksp, the solution is undersaturated, indicating no precipitate formation.

  • When Q = Ksp, the system is at equilibrium, maintaining a dynamic balance of dissolving and reforming ions.

  • If Q > Ksp, the solution is supersaturated, and excess ions begin to form a precipitate.
Understanding the role and calculation of the ion product is essential, as it provides insight into whether a solid will form from a given set of ion concentrations, allowing chemists to control and predict chemical processes in solution.
Solubility Product Constant (Ksp)
The solubility product constant (Ksp) is a unique value for each sparingly soluble ionic compound. It indicates the maximum concentration product of its ions that can dissolve in solution before the solid form begins to precipitate. Ksp is derived from the equilibrium condition of the dissolution of an ionic compound and reflects the degree of solubility of that substance in a given solvent.
To use Ksp efficiently, it is essential to recognize that larger Ksp values typically indicate more soluble compounds, while smaller values correspond to less soluble ones. Chemists utilize Ksp to evaluate the likelihood of precipitation in various reaction scenarios and also to help regulate processes such as the formation of scale in pipes or purifying processes in industrial applications.
By understanding and utilizing Ksp, you gain the ability to predict and manipulate solubility equilibria, empowering informed decision-making in both theoretical and applied chemistry contexts. It also allows chemists to optimize reactions for their desired outcomes, ensuring control over the occurrence or prevention of precipitation reactions.

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Most popular questions from this chapter

A sample of \(7.5 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) gas at \(22{ }^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 - \(\mathrm{L}\) solution of \(0.40 \mathrm{M} \mathrm{HCl}\). Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

A buffer is prepared by adding \(10.0 \mathrm{~g}\) of ammonium chloride \(\left(\mathrm{NH}_{4} \mathrm{Cl}\right)\) to \(250 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NH}_{3}\) solution. (a) What is the \(\mathrm{pH}\) of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

Use information from Appendix \(D\) to calculate the pH of (a) a solution that is \(0.250 \mathrm{M}\) in sodium formate \((\mathrm{HCOONa})\) and \(0.100 M\) in formic acid \((\mathrm{HCOOH}) ;\) (b) a solution that is \(0.510 \mathrm{M}\) in pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) and \(0.450 \mathrm{M}\) in pyridinium chloride \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right) ;\) (c) a solution that is made by combining \(55 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) hydrofluoric acid with \(125 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium fluoride.

(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 \(50.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaCHO}_{2}\)
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