Question

Barium is a toxic substance that can seriously impair heart function. For an \(X\) ray of the gastrointestinal tract, a patient drinks an aqueous suspension of \(20 \mathrm{~g} \mathrm{BaSO}_{4}\). If this substance were to equilibrate with the \(5.0 \mathrm{~L}\) of the blood in the patient's body, what would be \(\left[\mathrm{Ba}^{2+}\right] ?\) For a good estimate, we may assume that the \(K_{\mathrm{sp}}\) of \(\mathrm{BaSO}_{4}\) at body temperature is the same as at \(25^{\circ} \mathrm{C}\). Why is \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) not chosen for this procedure?

Short answer

[Ba^2+] = 1.05 x 10^{-5} M. Ba(NO3)2 is not used because it is much more soluble than BaSO4, releasing toxic levels of Ba^2+.

Step-by-step solution

Understand the Problem

We need to find the equilibrium concentration of \( \text{Ba}^{2+} \) ions in blood when \( 20 \text{ g} \) of \( \text{BaSO}_4 \) is added. Assume that all \( \text{BaSO}_4 \) is suspended evenly in the blood.

Recall Solubility Product Constant

The solubility product constant \( K_{sp} \) describes how much \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \) dissolve in solution. For \( \text{BaSO}_4 \), the equilibrium can be expressed as: \[ \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \] with \( K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \).

Find the Value of K_sp

Know the value of \( K_{sp} \) for \( \text{BaSO}_4 \) at \( 25^{\circ} \text{C} \), which is approximately \( 1.1 \times 10^{-10} \text{ mol}^2/\text{L}^2 \).

Set Up the Equilibrium Expressions

Since \([\text{Ba}^{2+}] = [\text{SO}_4^{2-}]\) at equilibrium when no additional sulfates are present, let \( x \) be the solubility of \( \text{BaSO}_4 \). Thus, \( [\text{Ba}^{2+}] = x \) and \( [\text{SO}_4^{2-}] = x \).

Solve for Solubility

Set the \( K_{sp} \) equation: \[ K_{sp} = x \times x = x^2 \] Therefore, \( x = \sqrt{1.1 \times 10^{-10}} \approx 1.05 \times 10^{-5} \text{ M} \).

Answer why Ba(NO3)2 is not used

\( \text{Ba(NO}_3\text{)}_2 \) is much more soluble than \( \text{BaSO}_4 \), which means it would release significantly higher concentrations of \( \text{Ba}^{2+} \) ions that could be toxic.

Key concepts

Barium Sulfate Solubility

Barium sulfate, represented by the chemical formula \( \text{BaSO}_4 \), is known for its extremely low solubility in water. This is particularly advantageous in medical imaging, such as X-rays of the gastrointestinal tract, since it remains largely undissolved. Its solubility is governed by its solubility product constant \( K_{sp} \). Low solubility ensures that when barium sulfate is ingested, it won't dissolve significantly into ions such as \( \text{Ba}^{2+} \), minimizing absorption into the bloodstream and reducing potential toxicity. The balance between the solid and its dissolved ions is described by the equation:\[ \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]This equilibrium state is the basis for calculating the minimal amount of barium ions present in a solution, making barium sulfate safe for use in medical imaging.

Equilibrium Concentration

Equilibrium concentration describes the ratio of products to reactants in a chemical solution that has reached a state of balance. At equilibrium, the concentration of barium ions \( [\text{Ba}^{2+}] \) and sulfate ions \( [\text{SO}_4^{2-}] \) in a saturated solution of barium sulfate is equal. This is because at equilibrium, \( \text{BaSO}_4 \) dissociates such that \( [\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = x \), where \( x \) is the solubility of barium sulfate in moles per liter.In the context of barium sulfate, knowing the equilibrium concentration helps in understanding how much of the compound will actually dissolve and what potential hazards might arise from ion release into the body. In this specific scenario, it helps medical practitioners ensure the safety profile of barium sulfate as a contrast agent by keeping released barium ions well below toxic levels.

Ksp Calculation

The \( K_{sp} \), or solubility product constant, is crucial for calculating how much of a substance can dissolve in a solution before reaching equilibrium. For barium sulfate, the \( K_{sp} \) equation is:\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = x^2 \]Given \( K_{sp} \) for \( \text{BaSO}_4 \) as approximately \( 1.1 \times 10^{-10} \text{ mol}^2/\text{L}^2 \) at room temperature, solving for \( x \) gives the solubility: \[ x = \sqrt{1.1 \times 10^{-10}} \approx 1.05 \times 10^{-5} \text{ M} \]This numerical value of \( x \) indicates the maximum concentration of \( \text{Ba}^{2+} \) ions in the solution. Understanding \( K_{sp} \) is fundamental in general chemistry for predictions about precipitate formation and solution saturation, making it an essential part of chemical equilibrium studies.

Toxicology in Chemistry

Toxicology studies the adverse effects of substances on living organisms. In chemistry, it is crucial to understand the toxicity profile of chemical compounds, such as barium and its salts. Pure barium is highly toxic, particularly affecting heart and nervous systems by interfering with potassium channels and blocking muscle function.However, the use of \( \text{BaSO}_4 \) in medical procedures is justified due to its incredibly low solubility, which confines barium ions within the compound, minimizing systemic absorption and toxic risk. Contrarily, \( \text{Ba(NO}_3\text{)}_2 \) is not suitable for similar purposes as it is highly soluble, releasing a greater concentration of \( \text{Ba}^{2+} \) ions. Therefore, understanding the solubility and \( K_{sp} \) values of chemicals is essential in assessing and mitigating potential health risks in various applications, especially when used in close proximity to biological systems.