Question

A BaSO \(_{4}\) slurry is ingested before the gastrointestinal tract is x-rayed because it is opaque to x-rays and defines the contours of the tract. \(\mathrm{Ba}^{2+}\) ion is toxic, but the compound is nearly insoluble. If \(\Delta G^{\circ}\) at \(37^{\circ} \mathrm{C}\) (body temperature) is \(59.1 \mathrm{~kJ} / \mathrm{mol}\) for the dissolution, $$ \mathrm{BaSO}_{4}(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) $$ what is \(\left[\mathrm{Ba}^{2+}\right]\) in the intestinal tract? (Assume that the only source of \(\mathrm{SO}_{4}^{2-}\) is the ingested slurry.

Short answer

The concentration of \(\text{Ba}^{2+}\) in the intestinal tract is approximately \(1.02 \times 10^{-5} \text{M}\).

Step-by-step solution

Write the Expression for \Delta G^{\circ}

Given the dissolution reaction: \[ \text{BaSO}_{4}(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_{4}^{2-}(aq) \], and \(\Delta G^{\circ} = 59.1\, \text{kJ/mol} \), we can write the expression for \(\Delta G^{\circ}\) as \[ \Delta G^{\circ} = -RT\ln K_{sp} \].

Convert \Delta G^{\circ} to Joules

Convert \(\Delta G^{\circ}\) from kJ to J: \[ 59.1 \text{kJ/mol} \times 1000 \frac{J}{1 \text{kJ}} = 59100 \text{J/mol} \].

Solve for K_{sp}

Use the expression \[ \Delta G^{\circ} = -RT\ln K_{sp} \] to find \(K_{sp}\). \(R = 8.314\, \text{J/mol}\cdot\text{K}\), and the temperature \(T = 310\, \text{K} \): \[ 59100 = -8.314 \cdot 310 \cdot \ln K_{sp} \].

Calculate \ln K_{sp}

Rearrange to solve for \(\ln K_{sp}\): \[ \ln K_{sp} = -\frac{59100}{8.314 \cdot 310} \].

Compute the Value of \ln K_{sp}

Perform the calculation: \[ \ln K_{sp} = -22.994 \].

Convert \ln K_{sp} to \K_{sp}

Exponentiate both sides to solve for \(K_{sp}\): \[ K_{sp} = e^{-22.994} \approx 1.05 \times 10^{-10} \].

Write the Solubility Product Expression

For the dissolution of \(\text{BaSO}_{4}(s)\), the solubility product expression is: \[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_{4}^{2-}] \]. Assume \([\text{Ba}^{2+}] = [\text{SO}_{4}^{2-}]\).

Solve for \left[\text{Ba}^{2+}\right]

Since \(K_{sp} = [\text{Ba}^{2+}]^2\), solve for \([\text{Ba}^{2+}]\): \[ [\text{Ba}^{2+}] = \sqrt{1.05 \times 10^{-10}} \approx 1.02 \times 10^{-5} \text{M} \].

Key concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \(\Delta G\), is a key concept in thermodynamics. It signifies the maximum reversible work that a thermodynamic system can perform at a constant temperature and pressure. Lower values of \(\Delta G\) indicate a more spontaneous reaction. When \(\Delta G\) is negative, the process is spontaneous, whereas if \(\Delta G\) is positive, the process is non-spontaneous.

In our context of BaSO\(_{4}\) dissolution, \(\Delta G^{\circ}\) is given as 59.1 kJ/mol, which shows this reaction is non-spontaneous under standard conditions. We use the formula \(\Delta G^{\circ} = -RT\text{ln} K_{sp}\) to connect Gibbs Free Energy with the solubility product constant, giving insight into solubility behavior.

Solubility Product Constant

The Solubility Product Constant, \(\text{K}_{sp}\), is an equilibrium constant that applies to the dissolution of sparingly soluble salts. For the reaction \(\text{BaSO}_{4}(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_{4}^{2-}(aq)\) it represents the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation.

\(\text{K}_{sp}\) provides insight into how much of the solid will dissolve in water. In our dissolution reaction, \(\text{K}_{sp}\) for BaSO\(_{4}\) was calculated using the formula derived from Gibbs Free Energy:

\[59100 = -8.314*310*\text{ln}K_{sp}\rightarrow K_{sp}=1.05*10^{-10}.\]
This value is very small, indicating that BaSO\(_{4}\) is poorly soluble, which is why it is used safely in x-ray procedures.

Thermodynamics in Chemistry

Thermodynamics studies the principles governing the energy exchanges in chemical reactions. It considers changes in heat, work, and energy within a system. Three main principles, also known as laws, govern thermodynamics:

• First law: Energy cannot be created or destroyed, only transformed.
• Second law: The entropy (disorder) of an isolated system always increases.
• Third law: As the temperature approaches absolute zero, the entropy of a perfect crystal approaches zero.

In our example, thermodynamics helps understand the solubility of BaSO\(_{4}\). The change in Gibbs Free Energy (\(\Delta G\)) is crucial. It determines if the dissolution (and thus solubility) will occur spontaneously. Since \(\Delta G\) for BaSO\(_{4}\)'s dissolution is positive, thermodynamics tells us that under normal human body temperature (310 K) conditions, BaSO\(_{4}\) does not dissolve easily. This property makes it safe for use in medical imaging.