Question

A BaSO4 slurry is ingested before the gastrointestinal tract is x-rayed because it is opaque to x-rays and defines the contours of the tract. Ba2+ ion is toxic, but the compound is nearly insoluble. If ΔG o at 37°C (body temperature) is 59.1 kJ/mol for the process BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq) what is the [Ba2+] in the intestinal tract? (Assume that the only source of SO42− is the ingested slurry.)

Answer 1

Explanation:

The given reaction equation is as follows.

         $BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)$

The value of $\Delta G^{o}$ = 59.1 kJ/mol

We know that ,

            $\Delta G^{o} = -RT ln K_{sp}$

 or,       $ln K_{sp} = -(\frac{\Delta G^{o}}{RT})$

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

            = -22.93

or,      $K_{sp} = e^{-22.93}$

                      = $1.1 \times 10^{-10}$

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

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

                                        = $\sqrt{ 1.1 \times 10^{-10}}$

                                         = $1.05 \times 10^{-5} M$

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