High School

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. The Ba\(^{2+}\) ion is toxic, but the compound is nearly insoluble.

If \(\Delta G^\circ\) at 37°C (body temperature) is 59.1 kJ/mol for the process:

\[ \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]

what is the \([\text{Ba}^{2+}]\) in the intestinal tract? (Assume that the only source of \(\text{SO}_4^{2-}\) is the ingested slurry.)

Answer :

Explanation:

The given reaction equation is as follows.

[tex]BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)[/tex]

The value of [tex]\Delta G^{o}[/tex] = 59.1 kJ/mol

We know that
,

[tex]\Delta G^{o} = -RT ln K_{sp}[/tex]

or, [tex]ln K_{sp} = -(\frac{\Delta G^{o}}{RT})
[/tex]

= -(\frac{59.1 kJ/mol}{(8.314 \times 10^{-3} kJ/mol.K \times 310 K))}[/tex]

= -22.93

or, [tex]K_{sp} = e^{-22.93}[/tex]

= [tex]1.1 \times 10^{-10}[/tex]

[tex]K_{sp} = [Ba^{2+}][ SO_{4}^{2-}][/tex]

Therefore, [tex][Ba^{2+}] =\sqrt{K_{sp}}[/tex]

= [tex]\sqrt{ 1.1 \times 10^{-10}}[/tex]

= [tex]1.05 \times 10^{-5} M[/tex]

Therefore, we can conclude that the value of [tex][Ba^{2+}][/tex] in the intestinal tract is [tex]1.05 \times 10^{-5} M[/tex].

Final answer:

The concentration of Ba2+ (barium ions) in the intestinal tract following the ingestion of BaSO4 slurry can be calculated using the equilibrium constant (K) derived from the Gibbs free energy equation. After performing the calculations, [Ba2+] is found to be 3.45 x 10⁻⁶ M.

Explanation:

The question is essentially asking for the concentration of Ba2+ ions in the intestinal tract following the ingestion of BaSO4 slurry, given a ΔG° value of 59.1 kJ/mol at 37°C. In order to solve this, we need to consider the process of the reaction: BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq). From the Gibbs free energy equation, we can calculate the equilibrium constant (K) as: K = e^(-ΔG°/RT), where R is the gas constant (R = 8.314 J/molK) and T is the temperature in Kelvin (37°C = 310K).

Therefore, ΔG° = -RTlnK -> K=e^(-ΔG°/RT). We find K = e^(-(−59.1×10³J mol⁻¹)/((8.314 J K⁻¹ mol⁻¹)(310 K))), which gives K = 1.19 x 10⁻¹⁰. Since the equilibrium involves the dissolution of one BaSO4 molecule to give one Ba2+ ion and one SO42- ion, the equilibrium concentrations of Ba2+ and SO42- are equal. Hence, [Ba2+] = sqrt(K) = sqrt(1.19 x 10⁻¹⁰) = 3.45 x 10⁻⁶ M.

In conclusion, the concentration of Barium ions, Ba2+, in the intestinal tract following ingestion of BaSO4 slurry would be around 3.45 x 10⁻⁶M.

Learn more about Concentration of Barium Ions here:

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