Barium sulfate is a contrast agent for X-ray scans that are most often associated with the gastrointestinal tract. Calculate the mass of \(\mathrm{BaSO}_{4}\) that can dissolve in \(100.0 \mathrm{~mL}\) of solution. The \(K_{\text {sp }}\) value for \(\mathrm{BaSO}_{4}\) is \(1.5 \times 10^{-9} .\)

The balanced equation for the dissolution of \(\mathrm{BaSO}_4\) is: \[\mathrm{BaSO}_4 \rightleftharpoons \mathrm{Ba^{2+}} + \mathrm{SO_4^{2-}}\]

Let's assume x mol/L of \(\mathrm{BaSO}_4\) dissolves in the solution. Since the dissolution in the equation is a 1:1 ratio, the molar concentrations of \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) can also be expressed as x mol/L.

Using the \(K_{sp}\) expression and the value given, we can determine the solubility x: \[K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]\] \[1.5 \times 10^{-9} = (x)(x)\] \[x^2 = 1.5 \times 10^{-9}\] Now, we find the value of x: \[x = \sqrt{1.5 \times 10^{-9}}\] \[x \approx 1.2 \times 10^{-5}\] Thus, the solubility of \(\mathrm{BaSO}_4\) is \(1.2 \times 10^{-5}\) mol/L.

To find the mass of \(\mathrm{BaSO}_4\) that can dissolve in 100 mL (0.100 L) of solution, we can multiply the solubility in mol/L by the volume in liters and the molar mass of \(\mathrm{BaSO}_4\): Mass of \(\mathrm{BaSO}_4\) = (Solubility) × (Volume) × (Molar mass of \(\mathrm{BaSO}_4\)) The molar mass of \(\mathrm{BaSO}_4\) can be calculated as follows: \[\mathrm{Ba:} ~ 137.33~ g/mol\] \[\mathrm{S:} ~ 32.07 ~ g/mol\] \[\mathrm{O_{4}:} ~ 4 \times 16.00 ~ g/mol = 64.00 ~g/mol\] \[\mathrm{Molar~mass~of~BaSO}_4 = 137.33 + 32.07 + 64.00 = 233.40 ~g/mol\] Now, we can find the mass of \(\mathrm{BaSO}_4\) that dissolves: Mass of \(\mathrm{BaSO}_4\) = \((1.2 \times 10^{-5}~mol/L) \times (0.100~L) \times (233.40 ~g/mol)\) Mass of \(\mathrm{BaSO}_4\) \(\approx 2.8 \times 10^{-4}~g\) So, approximately \(2.8 \times 10^{-4}~g\) of \(\mathrm{BaSO}_4\) can dissolve in 100 mL of solution.