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.)

Explanation:

The given reaction equation is as follows.

The value of = 59.1 kJ/mol

We know that ,

 or,      

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

            = -22.93

or,      

                      =

Therefore,      

                                        =

                                         =

Therefore, we can conclude that the value of in the intestinal tract is .

answer; /// i do believe that the correct answer is (a) boiling

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.