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° at 37°C (body temperature) is 75.0 kJ/mol for the process

Baso (8) = Ba2+(aq) + so-(aq)

what is [Ba²⁺] in the intestinal tract (multiply the number you calculate by 10⁶ and enter that number to 2 decimal places into the field)? (Assume that the only source of SO₄²⁻ is the ingested slurry.)

Answer 1

answer : 0.00

Explanation

Given : $\Delta$ Go = 75.0 kJ/mol = 75000 J/mol

ln(Ksp) = -($\Delta$Go) / (R * T)

where R = constant = 8.314 J/mol-K

T = absolute temperature = (37 + 273) K = 310 K

Substituting the values,

ln(Ksp) = -(75000 J/mol) / [(8.314 J/mol-K) * (310 K)]

ln(Ksp) = -29.1

Ksp = e-29.1

Ksp = 2.30 x 10-13

BaSO4 Ksp = [Ba2+][SO42-]

2.30 x 10-13 = (x) * (x)

2.30 x 10-13 = x2

x = (2.30 x 10-13)1/2

x = 4.832 x 10-7 M

[Ba2+] = 4.832 x 10-7 M

calculated value is 4.832 x 10-7 M

Now we multiply this number by 106

answer = (calculated value) * (106)

answer = (4.832 x 10-7 M) * (106)

answer = 5.12 x 10-5 M

answer = 0.0000512 M