If a patient ingests 380 mL of a saturated barium sulfate solution, how much toxic Ba ion have they consumed?

Question

Barium sulfate, BaSO4 , is used in medical imaging of the gastro-intestinal tract because it is opaque to x-rays. A barium sulfate solution is ingested by patients, whose stomach and intestines can then be visualized via x-ray imaging.

The Ksp of BaSO4 is 1.1 * 10^-10

Answer 1

This is presumably an assignment problem, so I guess you must use the data you were given. So, by the definition of solubility product constant Ksp:

Ksp = [Ba++][SO4--] = 1.1 * 10^-10

The concentration of Ba++ would be the square root of Ksp, since [Ba++] = [SO4--]:

[Ba++] = 1.05 * 10^-5 moles/liter.

To get the amount of Ba++ consumed:

1.05 * 10^-5 moles/liter * 0.380 liter = 4.0 * 10^-6 moles Ba++

4.0 * 10^-6 moles Ba++ * 137.3 g/mole Ba++ = 5.0 * 10^-4 grams of Ba++

or half a milligram of Barium

Answer 2

Strictly, a barium sulphate suspension is consumed: Practically none of the barium sulphate actually dissolves in water, as it would kill the patient if it did. Presumably though, it is the tiny amount which does dissolve with which you are concerned:
Now, the molar mass of barium is 137.3 gmol-1
sulphur: 32.1 gmol-1
oxygen: 16 gmol-1
So, the molar mass BaSO4 is 137.3 + 32.1 + 4(16) =
233.4 gmol-1

So, the mass percent of BaSO4 attributable to barium is:
137.3/233.4 x 100% = 58.8%

You haven't stated a saturation concentration with units, but presumably you have one.
Wikipedia gives me a value of 0.00115 gdm-3. (g/L)
You have 380 mL = 0.38L, so, the mass of BaSO4 in this amount =
0.00115 x 0.38 = 0.00057 g.
58.8% of this is barium, so 58% x 0.00057 = 0.000335 g.