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Adding \(1.85 \mathrm{g} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to \(500.0 \mathrm{mL}\) of saturated aqueous \(\mathrm{BaSO}_{4}:\) (a) reduces \(\left[\mathrm{Ba}^{2+}\right] ;\) (b) reduces \(\left[\mathrm{SO}_{4}^{2-}\right]\); (c) increases the solubility of \(\mathrm{BaSO}_{4} ;\) (d) has no effect.

Short Answer

Expert verified
The correct answers are: (a) reduces \(Ba^{2+}\), (b) reduces \(SO_4^{2-}\), (c) does not increase the solubility of BaSO4, and (d) does have an effect.

Step by step solution

01

Understanding the Common Ion Effect

When Na2SO4 is added to an aqueous solution of BaSO4, it dissociates into 2Na+ and \(SO_4^{2-}\) ions. This increases the \(SO_4^{2-}\) ion concentration in the solution. According to the common ion effect, the increase in the \(SO_4^{2-}\) concentration will shift the equilibrium to the left, reducing the \(Ba^{2+}\) concentration and hence the solubility of BaSO4.
02

Assess the change in ion concentration

Due to the shift of equilibrium, the \(Ba^{2+}\) & \(SO_4^{2-}\) ion concentrations both decrease. Hence, addition of Na2SO4 (a) reduces \(Ba^{2+}\) and (b) reduces \(SO_4^{2-}\).
03

Evaluate the effect on solubility

Regarding part (c), because the precipitation of BaSO4 is favored by the increased \(SO_4^{2-}\) ion concentration, the solubility of BaSO4 decreases, not increases. Therefore, (c) is not correct.
04

Summarize the overall effect

The overall effect of adding Na2SO4 to the saturated aqueous solution of BaSO4 is to reduce the concentrations of \(Ba^{2+}\) and \(SO_4^{2-}\) ions and decrease the solubility of BaSO4. Therefore, (d) is also not correct, as the addition of Na2SO4 does have an effect.

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Key Concepts

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

Solubility Product
The solubility product, often represented as \( K_{sp} \), is a measure of the solubility of a sparingly soluble salt in a solution. It defines the level at which a compound can dissolve in water to form a saturated solution. For the compound barium sulfate \( \text{(BaSO}_4) \), the solubility product can be written as:
  • \[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]
A saturated solution contains the maximum amounts of these ions, beyond which the compound will begin to precipitate, forming a solid. The solubility product helps us understand whether a precipitate will form when the concentrations of ions from added compounds reach or exceed this equilibrium constant.
In this exercise, adding sodium sulfate \((\text{Na}_2\text{SO}_4)\) increases the concentration of \([SO_4^{2-}]\), affecting the solubility equilibrium of \(\text{BaSO}_4\).
When the product of the concentrations \([Ba^{2+}][SO_4^{2-}]\) reaches or exceeds \( K_{sp}\), \(\text{BaSO}_4\) will precipitate, highlighting the principles of the solubility product.
Equilibrium Shift
Whenever an additional amount of an ion already part of the equilibrium system is added, the system will adjust to reestablish equilibrium. This is known as Le Chatelier's Principle, which states that if a change is imposed on a system at equilibrium, the system will respond to counteract the change and reestablish equilibrium.
In the case of adding \(\text{Na}_2\text{SO}_4\) to a saturated \(\text{BaSO}_4\) solution, the increase in \([SO_4^{2-}]\) ions causes the equilibrium of the \(\text{BaSO}_4\) dissolution to shift to the left. This shift results in more \(\text{BaSO}_4\) precipitating out of the solution, as seen in the equilibrium reaction:
  • \[ \text{BaSO}_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq) \]
With the increase in \([SO_4^{2-}]\), the system reduces the \([Ba^{2+}]\) ions to maintain the equilibrium. Consequently, fewer ions remain in solution, leading to reduced solubility of \(\text{BaSO}_4\). This illustrates how equilibrium shifts can dramatically impact ion concentrations and solubility.
Ion Concentration
Ion concentration refers to the amount of a particular ion present in a solution. This concentration affects the chemical behavior and properties of the solution, such as solubility, electrical conductivity, and reactivity. When compounds are dissolved in water, they dissociate into their constituent ions, which interact with other ions or molecules in the solution.
For this exercise, the common ion effect illustrates how ion concentration impacts a solution's equilibrium. By adding \(\text{Na}_2\text{SO}_4\), the concentration of \([SO_4^{2-}]\) increases. Since both \(\text{Na}_2\text{SO}_4\) and \(\text{BaSO}_4\) contribute \([SO_4^{2-}]\) ions, a rise in \([SO_4^{2-}]\) concentration due to added \(\text{Na}_2\text{SO}_4\) shifts the dissolution equilibrium to the left, causing \([Ba^{2+}]\) and \([SO_4^{2-}]\) ion concentrations to actually decrease as precipitation occurs.
Understanding how ion concentration influences chemical equilibrium is crucial for predicting solubility behavior and manipulating chemical reactions, demonstrating the intricate balance that governs saturated solutions and solubility.

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Most popular questions from this chapter

Determine the molar solubility of lead(II) azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2},\) in a buffer solution with \(\mathrm{pH}=3.00,\) given that \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=2.5 \times 10^{-9}\) \(\mathrm{HN}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{a}}=1.9 \times 10^{-5}\)

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.

A mixture of \(\mathrm{PbSO}_{4}(\mathrm{s})\) and \(\mathrm{PbS}_{2} \mathrm{O}_{3}(\mathrm{s})\) is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is \(\left[\mathrm{Pb}^{2+}\right]\) in the saturated solution? For \(\mathrm{PbSO}_{4}, K_{\mathrm{sp}}=1.6 \times 10^{-8} ;\) for \(\mathrm{PbS}_{2} \mathrm{O}_{3}, K_{\mathrm{sp}}=4.0 \times 10^{-7}\).

Write net ionic equations for each of the following observations. (a) When concentrated \(\mathrm{CaCl}_{2}(\mathrm{aq})\) is added to \(\mathrm{Na}_{2} \mathrm{HPO}_{4}(\mathrm{aq}),\) a white precipitate forms that is \(38.7 \%\) Ca by mass. (b) When a piece of dry ice, \(\mathrm{CO}_{2}(\mathrm{s}),\) is placed in a clear dilute solution of limewater \(\left[\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})\right]\), bubbles of gas evolve. At first, a white precipitate forms, but then it redissolves.

A handbook lists the \(K_{\mathrm{sp}}\) values \(1.1 \times 10^{-10}\) for \(\mathrm{BaSO}_{4}\) and \(5.1 \times 10^{-9}\) for \(\mathrm{BaCO}_{3} .\) When saturated \(\mathrm{BaSO}_{4}(\mathrm{aq})\) is also made with \(0.50 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}),\) a precipitate of \(\mathrm{BaCO}_{3}(\mathrm{s})\) forms. How do you account for this fact, given that \(\mathrm{BaCO}_{3}\) has a larger \(K_{\mathrm{sp}}\) than does \(\mathrm{BaSO}_{4} ?\)

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