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Chapter 17: Problem 87

Using the following data, calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), one of the least soluble of the common nitrate salts. $$ \begin{array}{lc} \text { Species } & \Delta G_{\mathrm{f}}^{\circ} \\ \mathrm{Ba}^{2+}(a q) & -561 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{NO}_{3}^{-}(a q) & -109 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(s) & -797 \mathrm{~kJ} / \mathrm{mol} \\ \hline \end{array} $$

Short Answer

Expert verified
The solubility product constant, \(K_{sp}\), for the slightly soluble salt \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) is approximately \(1554.1\).

Step by step solution

01

Write the balanced dissolution reaction

Write the balanced equation for the dissolution of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) in water: \[ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(s) \leftrightharpoons \mathrm{Ba}^{2+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) \]
02

Calculate the standard Gibbs free energy change for the reaction

Using the given standard Gibbs free energy of formation values, we can calculate the standard Gibbs free energy change, \(\Delta G^\circ_{rxn}\), for the reaction. The formula for this is: \[ \Delta G^\circ_{rxn} = \sum_{products} n_i \Delta G_{f,i}^{\circ} - \sum_{reactants} n_j \Delta G_{f,j}^{\circ} \] Plugging in the values for the dissolution reaction, we have: \[ \Delta G^\circ_{rxn} = \left[\left(1\right) \left(-561~\mathrm{kJ/mol}\right) + \left(2\right) \left(-109~\mathrm{kJ/mol}\right)\right] - \left[\left(1\right) \left(-797~\mathrm{kJ/mol}\right)\right] \]
03

Calculate the standard Gibbs free energy change value

Solving the expression from the previous step, we find the standard Gibbs free energy change for the reaction: \[ \Delta G^\circ_{rxn} = \left(-561 + 2 \times -109\right) - \left(-797\right) \\ \Delta G^\circ_{rxn} = \left(-561 - 218\right) + 797 \\ \Delta G^\circ_{rxn} = -779 + 797 \\ \Delta G^\circ_{rxn} = 18~\mathrm{kJ/mol} \]
04

Use the standard Gibbs free energy change to find the equilibrium constant

The relationship between standard Gibbs free energy change, the equilibrium constant, and temperature is given by the formula: \[ \Delta G^\circ_{rxn} = -RT \ln{K} \] where \(R\) is the gas constant (8.314 J/mol•K) and \(T\) is the temperature in Kelvin. Assuming the reaction takes place at room temperature, \(25^\circ\)C, which is equal to \(298.15\) K, we can solve for the equilibrium constant, \(K\): \[ 18 \times 10^{3} = -\left(8.314\right) \left(298.15\right) \ln{K} \\ \]
05

Calculate the equilibrium constant

Solving for \(K\) in the expression from the previous step, we get: \[ \ln{K} = \dfrac{-18 \times 10^{3}}{-(8.314)(298.15)} \\ \ln{K} = 7.35 \] Find the value of \(K\) by applying the exponential function: \[ K = e^{7.35} \\ K = 1554.1 \]
06

Solving for \(K_{\mathrm{sp}}\)

Since the equilibrium constant \(K\) represents the solubility product constant, \(K_{sp}\), for \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), we have: \[ K_{\mathrm{sp}} \approx 1554.1 \] In conclusion, the value of the solubility product constant, \(K_{sp}\), for the slightly soluble salt \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) is approximately \(1554.1\).

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Most popular questions from this chapter

Consider the following reaction at \(800 . \mathrm{K}\) : $$ \mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g) $$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm}\). Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(\mathrm{g})\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

Given the following data: $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) & \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g) & & \Delta G^{\circ}=-394 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l)\).

Consider the reactions $$ \begin{aligned} \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q) \end{aligned} $$ where $$ \text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} $$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text {reaction } 2}>K_{\text {reaction } 1} .\) Explain.

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?
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