Questions: Chapter 15 Homework 22. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 15.4). This use of BaSO4 is possible because of its low solubility. Calculate the molar solubility of BaSO4 and the mass of barium present in 1.00 L of water saturated with BaSO4.

Solution Steps

The solubility product constant (\(K_{sp}\)) for barium sulfate (\(BaSO_4\)) is given by the expression: \[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]

Let \(s\) be the molar solubility of \(BaSO_4\). This means that in a saturated solution, the concentration of \(Ba^{2+}\) ions and \(SO_4^{2-}\) ions will both be \(s\).

Since the concentrations of \(Ba^{2+}\) and \(SO_4^{2-}\) are both \(s\), we can substitute \(s\) into the \(K_{sp}\) expression: \[ K_{sp} = s \cdot s = s^2 \]

The \(K_{sp}\) value for \(BaSO_4\) is \(1.1 \times 10^{-10}\). Therefore, we can solve for \(s\): \[ s^2 = 1.1 \times 10^{-10} \] \[ s = \sqrt{1.1 \times 10^{-10}} \] \[ s = 1.0488 \times 10^{-5} \]

The molar mass of barium (\(Ba\)) is 137.33 g/mol. The mass of barium in 1.00 L of water saturated with \(BaSO_4\) is: \[ \text{mass of } Ba = s \times \text{molar mass of } Ba \] \[ \text{mass of } Ba = 1.0488 \times 10^{-5} \, \text{mol/L} \times 137.33 \, \text{g/mol} \] \[ \text{mass of } Ba = 0.001440 \, \text{g} \]

Final Answer