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A solution is \(1.8 \times 10^{-4} M \mathrm{Co}^{2+}\) and \(0.20 \mathrm{M} \mathrm{HSO}_{4}\) The solution also contains \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). What should be the minimum molarity of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) to prevent the precipitation of cobalt(II) sulfide when the solution is saturated with hydrogen sulfide \(\left(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\right)\) ?

Short Answer

Expert verified
The minimum Na2SO4 molarity is governed by system equilibrium, roughly balancing near 0.10M while avoiding CoS precipitation by manipulating [S^{2-}].

Step by step solution

01

Determine Solubility Product Expression

For cobalt(II) sulfide, CoS, the solubility product expression is given by \( K_{sp} = [Co^{2+}][S^{2-}] \). We need to ensure that the conditions prevent \( [Co^{2+}] \cdot [S^{2-}] \leq K_{sp}(CoS) \).
02

Find Required Ionic Conditions

Given \( [Co^{2+}] = 1.8 \times 10^{-4} \, M \), we need to calculate the concentration of \( S^{2-} \), which depends on the dissociation equilibrium of \( H_2S \): \( H_2S \rightleftharpoons 2H^+ + S^{2-} \). However, since we are adding Na2SO4, the reaction shifts toward the left, reducing \( [S^{2-}] \).
03

Calculate the Influence of \\([H^+]\\)

The sulfur ion exists in multiple forms: \( H_2S, HS^-, S^{2-} \). Since we're interested in precipitation of \( S^{2-} \) from \( H_2S \), manage with the equilibrium \( HSO_4^- \rightleftharpoons H^+ + SO_4^{2-} \) and the presence of Na2SO4 providing additional \( SO_4^{2-} \).
04

Eliminate Variables Through Mass Balance

Consider the added pH influence, where \( HSO_4^- \) influences \( [H^+] \). As we saturate with Na2SO4, less \( H^+ \) is available in solution equilibria, allowing controlled \( [S^{2-}] \) to prevent CoS ppt.
05

Perform Stoichiometry for Final Conditions

For preventing CoS ppt, the saturation multiplying available Na2+ controls sulfate. Using the minimal \( Na_2SO_4 \) concentration aligns \([SO_4^{2-}]\), minimizing \( [S^{2-}] \). Thusly, balancing occurs at PULL theoretical saturation point against \([Co^{2+}]\), since \( [S^{2-}] \) \( \propto \) \( 1/[SO_4^{2-}]\).
06

Approximate Minimum Molarity Requirement

By calculations with precipitative predicate control, empirical molarity ranges go \( 1.8 \times 10^{-4} \), reducing \( S^{2-} \) below potential product formation. Accounting solvable endpoint derives at a theoretical \( Min[Na_2SO_4] \) aligning to \( 0.10M \) H2S potential into stoichiometry balance.

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Key Concepts

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

Cobalt(II) Sulfide
Cobalt(II) Sulfide, or CoS, is an inorganic compound and is considered a sparingly soluble salt. This means that it does not dissolve readily in water. The solubility product constant, denoted as \( K_{sp} \), for CoS quantifies its solubility. The smaller this value, the less soluble the substance is. In the context of precipitation chemistry, controlling the solubility of CoS is key to preventing its formation in a given solution. To prevent CoS from precipitating, conditions must satisfy \([Co^{2+}][S^{2-}] \leq K_{sp}(CoS) \). Understanding how to alter these concentrations in solution is critical for controlling the chemical behavior of cobalt sulfide.
Precipitation Reaction
Precipitation reactions involve the formation of a solid, or precipitate, from an aqueous solution. In this exercise, we are concerned with preventing the formation of solid CoS. The reaction occurs when two ions in solution combine to form an insoluble compound that eventually comes out of the solution as a solid. The prevention of precipitation is managed by manipulating the concentrations of the reactant ions. For the precipitation of cobalt(II) sulfide, this involves managing the concentration of \( S^{2-} \), which in turn depends on the concentration of hydrogen sulfide \( H_2S \) and other ions like \( SO_4^{2-} \) present in the solution. By understanding the equilibrium between these ions, chemists can prevent unwanted precipitation.
Ionic Equilibrium
Ionic equilibrium is the balance that exists when a reversible reaction occurs between ions in a solution. In the context of this exercise, the focus is on the equilibrium involving sulfur ions derived from \( H_2S \). The dissociation of \( H_2S \) into \( H^+ \) and \( S^{2-} \) ions is described by its equilibrium process: \( H_2S \rightleftharpoons 2H^+ + S^{2-} \). The presence of \( SO_4^{2-} \), provided by \( Na_2SO_4 \), plays a significant role in shifting this equilibrium. With the addition of \( Na_2SO_4 \), the equilibrium shifts to reduce the concentration of \( S^{2-} \), thus controlling the precipitation of \( CoS \). By understanding and manipulating ionic equilibria, chemists can control reactions in a solution.
Hydrogen Sulfide Saturation
Hydrogen sulfide (\( H_2S \)) in solution can dissociate to produce \( S^{2-} \), contributing to the potential precipitation of cobalt sulfide. The term 'saturation' with \( H_2S \) refers to reaching a concentration at which no more can be dissolved in the solution. In this particular scenario, the solution is saturated with \( 0.10 \, M \) \( H_2S \). Saturation influences the amount of \( S^{2-} \) available, thus affecting the solubility equilibrium of CoS. The presence of other ions, such as \( SO_4^{2-} \) from \( Na_2SO_4 \), in the solution further influences this saturation level. By adjusting these factors appropriately, the formation of \( S^{2-} \) can be minimized, which is key to preventing the precipitation of \( CoS \). Chemistry students must understand these interactions to control precipitative outcomes in laboratory settings.

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Most popular questions from this chapter

Gout is a painful inflammation of the joints caused by an excess of uric acid in the blood and its precipitation as sodium urate (the sodium salt of uric acid), \(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{~N}_{4} \mathrm{O}_{3},\) in tissues of the joints. The solubility of sodium urate in aqueous solution is \(7.0 \mathrm{mg}\) per \(100 \mathrm{~mL}\) at body temperature, \(37^{\circ} \mathrm{C}\). The normal level of sodium ion in blood plasma is \(3.2 \mathrm{~g}\) per liter. What would be the concentration of urate ion in blood plasma if it were saturated with sodium urate?

Write solubility product expressions for the following compounds. a) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) b) \(\mathrm{FePO}_{4}\) c) \(\mathrm{Ag}_{2} \mathrm{~S}\) d) \(\mathrm{PbI}_{2}\)

Consider three hypothetical ionic solids: \(\mathrm{AX}, \mathrm{AX}_{2}\), and \(\mathrm{AX}_{3}\) (each \(\mathrm{X}\) forms \(\mathrm{X}^{-}\) ). Each of these solids has the same \(K_{s p}\) value, \(5.5 \times 10^{-7}\). You place 0.25 mol of each compound in a separate container and add enough water to bring the volume to \(1.0 \mathrm{~L}\) in each case. a. Write the chemical equation for each of the solids dissolving in water. b. Would you expect the concentration of each solution to be \(0.25 M\) in the compound? Explain, in some detail, why or why not. c) Would you expect the concentrations of the A cations \(\left(\mathrm{A}^{+}, \mathrm{A}^{2+},\right.\) and \(\left.\mathrm{A}^{3+}\right)\) in the three solutions to be the same? Does just knowing the stoichiometry of each reaction help you determine the answer, or do you need something else? Explain your answer in detail, but without doing any arithmetic calculations. d. Of the three solids, which one would you expect to have the greatest molar solubility? Explain in detail, but without doing any arithmetic calculations. e. Calculate the molar solubility of each compound.

The solubility of magnesium oxalate, \(\mathrm{MgC}_{2} \mathrm{O}_{4}\), in water is \(0.0093 \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{s p}\).

Write the solubility product expression for the salt $$ \begin{array}{l} \mathrm{Ag}_{3} \mathrm{PO}_{4} \text { . } \\ \text { a) } K_{s p}=3\left[\mathrm{Ag}^{+}\right] \times\left[\mathrm{PO}_{4}^{3-}\right] \\ \text { b) } K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{PO}_{4}^{3-}\right]^{3} \\ \text { c) } K_{s p}=\left[\mathrm{Ag}^{+}\right]^{3}\left[\mathrm{P}^{3-}\right]\left[\mathrm{O}^{2-}\right]^{4} \\\ \text { d) } K_{s p}=\left[\mathrm{Ag}^{3+}\right]\left[\mathrm{PO}_{4}^{-}\right]^{3} \\ \text { e } K_{s p}=\left[\mathrm{Ag}^{+}\right]^{3}\left[\mathrm{PO}_{4}^{3-}\right] \end{array} $$

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