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Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C},\) the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess CaSO \(_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{L},\) what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

Short Answer

Expert verified
(a) The equilibrium concentrations of Ca虏鈦 and SO鈧劼测伝 are \(4.9 \times 10^{-3}\) mol/L. (b) The minimum mass of CaSO鈧 needed to achieve equilibrium in 1.4 L of solution is 0.933 g.

Step by step solution

01

Write the expression for the equilibrium constant

For the given reaction: \[\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\] The equilibrium constant expression is given by: \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{CaSO}_{4}]}\] Since the concentration of CaSO鈧(s) is considered constant, we can write it as: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\] Given, \(K_c = 2.4 \times 10^{-5}\)
02

Calculate the equilibrium concentrations

Let [Ca虏鈦篯=[SO鈧劼测伝]=x mol/L, at equilibrium since in the balanced reaction, for every one mole of CaSO鈧 that dissolves, one mole of Ca虏鈦 and one mole of SO鈧劼测伝 ions are produced. So we can write: \[K_c = x^2\] Now, we can solve for x: \[2.4 \times 10^{-5} = x^2\] \[x = \sqrt{2.4 \times 10^{-5}}\] \[x = 4.9 \times 10^{-3}\] Hence, the equilibrium concentrations of Ca虏鈦 and SO鈧劼测伝 are: \[[\mathrm{Ca}^{2+}] = [\mathrm{SO}_{4}^{2-}] = 4.9 \times 10^{-3}\,\text{mol/L}\]
03

Determine the minimum mass of CaSO鈧

Given the volume of the solution is 1.4 L, we can find the moles of Ca虏鈦 and SO鈧劼测伝 ions in the solution: Moles of Ca虏鈦 = Moles of SO鈧劼测伝 = \([\mathrm{Ca}^{2+}] \times \text{Volume}\) \[= (4.9 \times 10^{-3}\,\text{mol/L}) \times 1.4\,\text{L}\] \[= 6.86 \times 10^{-3}\,\text{mol}\] Now, we need to convert the moles of Ca虏鈦 into grams of CaSO鈧. The molar mass of CaSO鈧: Molar mass = 40.08 (Ca) + 32.07 (S) + 4*16 (O) = 136.14 g/mol Minimum mass of CaSO鈧 = moles of Ca虏鈦 * molar mass of CaSO鈧 \[= 6.86 \times 10^{-3}\,\text{mol} \times 136.14\,\text{g/mol}\] \[= 0.933\,\text{g}\] #Answer# (a) The equilibrium concentrations of Ca虏鈦 and SO鈧劼测伝 are 4.9 x 10鈦宦 mol/L. (b) The minimum mass of CaSO鈧 needed to achieve equilibrium in 1.4 L of solution is 0.933 g.

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Key Concepts

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

Equilibrium Constant
Understanding the equilibrium constant is essential when studying chemical equilibria. It is a numerical value characterized by the ratio of the concentrations of the products to the reactants for a reversible reaction at equilibrium. The equation for the equilibrium constant, denoted as \( K_c \) for reactions in solution, is given by following formula:
  • \( K_c = \frac{[products]}{[reactants]} \)
The reaction for calcium sulfate dissolution is:\[ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \]Here, the equilibrium constant is \( K_c = 2.4 \times 10^{-5} \), which suggests that at equilibrium, the product concentrations are relatively low, indicating low solubility. This low value indicates that only a small amount of calcium sulfate dissolves in water. The equilibrium constant is not changed by the initial amount of solid or changes in total pressure, but it will vary with temperature. Knowing \( K_c \) allows us to predict the concentrations of ions at equilibrium, which is incredibly useful for determining how soluble a substance is in a solution.
Saturated Solution
A saturated solution is one where the maximum amount of solute has been dissolved in the solvent, beyond which no more solute can dissolve at a given temperature. When mixing calcium sulfate with water until no more dissolves, you obtain a saturated solution. This is a state where the rate at which the solute dissolves equals the rate at which it precipitates from the solution.

In the context of the given reaction, mixing an excess of calcium sulfate with water results in the establishment of an equilibrium between the dissolved ions and the undissolved solid. At 25掳C, once equilibrium is reached, any additional calcium sulfate added will remain undissolved because the solution has achieved its maximum solubility for the temperature. The concentrations of \( \text{Ca}^{2+} \) and \( \text{SO}_{4}^{2-} \) are determined by the equilibrium constant \( K_c \) as demonstrated in the solution steps, reflecting the whole saturated solution.
Ionic Concentration
The concept of ionic concentration describes the amount of ions present in a solution. In a saturated solution of calcium sulfate, the dissolved ions \( \mathrm{Ca}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) are in equilibrium. The concentration of these ions at equilibrium can be found using the equilibrium constant. For calcium sulfate in water:
  • The concentration of \( \text{Ca}^{2+} \) is the same as \( \text{SO}_{4}^{2-} \), due to their 1:1 stoichiometry in the dissociation reaction.
  • This concentration is calculated as \( \sqrt{K_c} \) since the dissolved ions contribute equally to the product term \( [\text{Ca}^{2+}][\text{SO}_{4}^{2-}] \).
Upon solving \( K_c = x^2 \) with \( x = 4.9 \times 10^{-3} \text{ mol/L} \), we find that both ion concentrations are equal and reflect the saturation point of the solution. Understanding ionic concentration helps predict the strength and properties of the solution, impacting reactions, solubility, and conductivity.
Solubility Product
The solubility product, often denoted as \( K_{sp} \), is an application of the equilibrium constant specifically for dissolving salts. It's useful in quantifying a salt's solubility in solution. For our reaction:\[ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \]The solubility product is the product of the molar concentrations of the ions, each raised to the power of their stoichiometric coefficients, and is given by:
  • \( K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] \)
For calcium sulfate, \( K_{sp} = 2.4 \times 10^{-5} \), effectively quantifying its solubility in terms of products of ion concentrations at equilibrium. The solubility product is crucial in predicting when a salt will precipitate from a solution or dissolve into ions. If the product of the ion concentrations exceeds \( K_{sp} \), the solution is supersaturated, and precipitation occurs. Conversely, if below \( K_{sp} \), the solution is unsaturated, and additional solid can dissolve. By understanding this concept, chemists can manipulate conditions to control salt solubility and precipitation.

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Most popular questions from this chapter

For the equilibrium $$\mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)$$ \(K_{p}=0.052\) at \(60^{\circ} \mathrm{C}\) (a) Calculate \(K_{C}\) (b) After 3.00 \(\mathrm{g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed 1.500 -L. vessel at \(60^{\circ} \mathrm{C}\) , the vessel is charged with 0.0500 \(\mathrm{g}\) of \(\mathrm{BCl}_{3}(g) .\) What is the equilibrium concentration of \(\mathrm{PH}_{3} ?\)

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at 500 \(\mathrm{K}\) . An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\) , respectively. (a) Calculate \(K_{p}\) for this reaction at 500.0 \(\mathrm{K}\) . (b) If the vessel has a volume of 5.00 \(\mathrm{L}\) , calculate \(K_{c}\) at this temperature.

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

At \(700 \mathrm{K},\) the equilibrium constant for the reaction $$\operatorname{CCI}_{4}(g) \Longrightarrow \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=0.76 .\) A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\) ,which then reaches equilibrium at 700 \(\mathrm{K}\) . (a) What fraction of the CCl\(_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?(\mathbf{b})\) what are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

The equilibrium constant \(K_{c}\) for \(C(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) 2 \(\mathrm{CO}(g)\) is 1.9 at 1000 \(\mathrm{K}\) and 0.133 at 298 \(\mathrm{K}\) . (a) If excess\(\mathrm{C}\) is allowed to react with 25.0 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) in a 3.00 -L vessel at \(1000 \mathrm{K},\) how many grams of CO are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

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