/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Consider the reaction $$\mathr... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 55

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 \(\mathrm{CaSO}_{4}(s)\) 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 \(\mathrm{SO}_{4}^{2-} ?(\mathbf{b})\) 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
The equilibrium concentrations of Ca虏鈦� and SO鈧劼测伝 are both \(4.9 \times 10^{-3} \, \text{mol/L}\). The minimum mass of CaSO鈧� needed to achieve equilibrium in a 1.4 L solution is approximately 0.934 g.

Step by step solution

01

Write down the given information.

We are given the following information: - Reaction: \(\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \) - Equilibrium constant, \(K_{c} = 2.4 \times 10^{-5}\) at 25掳C - Volume of the solution, \(V = 1.4\) L
02

Define the relationships between equilibrium constant and reactant and product concentrations.

According to the reaction, the equilibrium constant can be written as: $$K_{c} = \frac{[\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{CaSO}_{4}]}$$ Since CaSO鈧� is a solid, we can rewrite the equation as: $$K_{c} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]$$
03

Determine the equilibrium concentrations of Ca虏鈦� and SO鈧劼测伝.

Let the equilibrium concentration of Ca虏鈦� be x mol/L. Since the stoichiometry of the reaction is 1:1, the equilibrium concentration of SO鈧劼测伝 will also be x mol/L. Now, we can substitute their concentrations into the equation for the equilibrium constant: $$K_{c} = x^2$$ $$2.4 \times 10^{-5} = x^2$$ Solve for x: $$x = \sqrt{2.4 \times 10^{-5}}$$ $$x = 4.9 \times 10^{-3} \, \text{mol/L}$$ So, the equilibrium concentrations of Ca虏鈦� and SO鈧劼测伝 are both \(4.9 \times 10^{-3} \, \text{mol/L}\).
04

Calculate the minimum mass of CaSO鈧� needed to achieve equilibrium.

We know that the equilibrium concentration of Ca虏鈦� is \(4.9 \times 10^{-3} \, \text{mol/L}\). As the stoichiometry of the reaction is 1:1, the equilibrium concentration of CaSO鈧� is the same. Since we know the volume of the solution (1.4 L), we can find the moles of CaSO鈧� required to reach equilibrium: Moles of CaSO鈧� = concentration 脳 volume Moles of CaSO鈧� = \(4.9 \times 10^{-3} \, \text{mol/L} \times 1.4 \, \text{L}\) Moles of CaSO鈧� = \(6.86 \times 10^{-3} \, \text{mol}\) Now, calculate the molar mass of CaSO鈧�: Molar mass of CaSO鈧� = 40.08 (Ca) + 32.07 (S) + 4 脳 16.00 (O) = 136.14 g/mol Finally, find the minimum mass of CaSO鈧� needed: Mass of CaSO鈧� = moles 脳 molar mass Mass of CaSO鈧� = \(6.86 \times 10^{-3} \, \text{mol} \times 136.14 \, \text{g/mol}\) Mass of CaSO鈧� = 0.934 g The minimum mass of CaSO鈧� needed to achieve equilibrium in a 1.4 L solution is approximately 0.934 g.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \(K_c\), is crucial in understanding chemical equilibrium. It helps us predict how a chemical reaction will behave over time, especially when it reaches a state called equilibrium. At equilibrium, the rates of the forward and backward reactions are equal.
For our reaction, \(\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\), the equilibrium constant expression is:
  • \(K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\)
Here, concentrations of solids like \(\mathrm{CaSO}_{4}\) don't appear in the \(K_c\) expression, as solids have constant concentrations. Also, \(K_c = 2.4 \times 10^{-5}\) at 25掳C indicates a low concentration of ions at equilibrium, signifying a sparingly soluble salt. Thus, knowing \(K_c\) allows us to estimate ion concentrations in a solution.
Solubility Product
The solubility product, often symbolized as \(K_{sp}\), relates to the solubility of ionic compounds like \(\mathrm{CaSO}_{4}\). It is a special kind of equilibrium constant that applies specifically to dissolution processes in water.
  • The expression for \(K_{sp}\) for \(\mathrm{CaSO}_{4}\) is similar to \(K_c\):
  • \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\)
Since the stoichiometry of this dissolution is 1:1 for \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\), we see that the concentrations of both ions are equal at equilibrium.
Thus, \(K_{sp} = x^2\), where \(x\) is the molar solubility of \(\mathrm{CaSO}_{4}\).
Solving \(x = \sqrt{2.4 \times 10^{-5}}\) gives \(x = 4.9 \times 10^{-3} \, \text{mol/L}\), showing how much \(\mathrm{CaSO}_{4}\) dissolves at equilibrium.
Concentration Calculations
Concentration calculations involve finding how much solute is dissolved in a given volume of solvent, which is vital for understanding equilibrium in reactions. In our example, we calculated the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ions:
  • Both are \(4.9 \times 10^{-3} \, \text{mol/L}\).
Next, to find the minimum mass of \(\mathrm{CaSO}_{4}\) needed:
  • Calculate moles: \(\text{Moles of } \mathrm{CaSO}_{4} = 4.9 \times 10^{-3} \, \text{mol/L} \times 1.4 \, \text{L} = 6.86 \times 10^{-3} \, \text{mol}\)
  • Determine molar mass: \(136.14 \, \text{g/mol}\).
  • Calculate mass: \(\text{Mass} = 6.86 \times 10^{-3} \, \text{mol} \times 136.14 \, \text{g/mol} = 0.934 \, \text{g}\).
Therefore, approximately 0.934 g of \(\mathrm{CaSO}_{4}\) is needed, showcasing how concentration calculations help us determine the amount of substance required to achieve equilibrium in a solution.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a \(2.00-\mathrm{L}\) vessel is found to contain \(0.0406 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH},\) \(0.170 \mathrm{~mol} \mathrm{CO},\) and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K} .\) Calculate \(K_{c}\) at this temperature.

A \(0.831-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 - \(\mathrm{L}\) container and heated to \(1100 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the total pressure in the container is \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\) at \(700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\). (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{~S} ?\) (c) Calculate the values of \(\mathrm{K}_{c}\) and \(K_{p}\) if you rewrote the balanced chemical equation with \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) instead of \(2 \mathrm{~mol}\).

In Section 11.5 we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for \(K_{p} .\) (b) By using data in Appendix \(\mathrm{B}\), give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C} .(\mathrm{c})\) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?
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