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Chapter 10: Problem 97

People who have the misfortune of going through a series of \(\mathrm{X}\) rays of the gastrointestinal tract are often given a suspension of solid barium sulfate in water to drink. \(\mathrm{BaSO}_{4}\) is used instead of other \(\mathrm{Ba}^{2+}\) salts, which also reflect \(\mathrm{X}\) rays, because it is relatively insoluble in water. (Thus the patient is exposed to the minimum amount of toxic \(\mathrm{Ba}^{2+}\) ion.) What is the solubility product for barium sulfate if 1 gram of \(\mathrm{BaSO}_{4}\) dissolves in 400,000 grams of water?

Short Answer

Expert verified
Ksp for BaSO鈧� is approximately 1.14 脳 10鈦宦光伆.

Step by step solution

01

Determine Solubility in Moles per Liter

First, calculate the molarity of soluble BaSO鈧�. We begin with the data that 1 gram of BaSO鈧� dissolves in 400,000 grams of water. The molar mass of BaSO鈧� is approximately 233.39 g/mol. Therefore, the number of moles of BaSO鈧� in 1 gram is:\[\frac{1\text{ g}}{233.39\text{ g/mol}} \approx 0.00428\text{ mol}\]Since 400,000 grams of water is equivalent to 400 liters (because 1 gram = 1 mL and 1,000 mL = 1 L), the molarity (concentration) of BaSO鈧� in moles per liter is:\[\frac{0.00428\text{ mol}}{400\text{ L}} \approx 1.07 \times 10^{-5}\text{ mol/L}\]
02

Write the Dissolution Reaction

Write the chemical equation for the dissolution of BaSO鈧� in water:\[\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)\]This equation indicates that 1 mole of solid BaSO鈧� produces 1 mole of Ba虏鈦� ions and 1 mole of SO鈧劼测伝 ions when it dissolves.
03

Express the Solubility Product (Ksp) Formula

The solubility product expression for this dissolution is:\[K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]\]Given the stoichiometry of the reaction, for every x moles of BaSO鈧� that dissolve, x moles of Ba虏鈦� and x moles of SO鈧劼测伝 are produced. Therefore, in terms of concentration:\[K_{sp} = x \times x = x^2\]
04

Calculate the Solubility Product (Ksp)

From Step 1, we found that the concentration of Ba虏鈦� ions, [Ba虏鈦篯, is approximately 1.07 脳 10鈦烩伒 mol/L. The concentration of SO鈧劼测伝 ions, [SO鈧劼测伝], is the same because of the 1:1 stoichiometry.Thus,\[K_{sp} = (1.07 \times 10^{-5})^2 = 1.14 \times 10^{-10}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Barium Sulfate
Barium sulfate ( BaSO鈧� ) is a white crystalline solid known for its very low solubility in water. This property makes it quite unique compared to other barium compounds. Because it is poorly soluble, it is often used medically for ingesting as a contrast agent during X-rays of the gastrointestinal tract. Unlike other barium salts, BaSO鈧� is not highly toxic, because only a minimal amount dissolves into the bloodstream. When exposed to X-rays, its density helps produce clear images by providing the necessary contrast. This distinct feature is why BaSO鈧� is chosen for this application, keeping patient safety in mind while offering superior imaging results.
Solubility Calculation
To find the solubility of a substance like barium sulfate, you will first need to determine how much of it dissolves in water to reach a saturated solution. In this instance, starting with 1 gram of barium sulfate in 400,000 grams of water, we convert grams to liters. Since 1 gram of water equals 1 milliliter, 400,000 grams converts to 400 liters. Knowing the molar mass of BaSO鈧� is approximately 233.39 g/mol, the solubility in moles per liter is calculated as follows:
  • Convert mass to moles: \(\frac{1\text{ g}}{233.39\text{ g/mol}} = 0.00428\text{ mol}\)
  • Convert volume to liters and find molarity: \(\frac{0.00428\text{ mol}}{400\text{ L}} = 1.07 \times 10^{-5}\text{ mol/L}\)
The very low molarity reflects the very limited amount of BaSO鈧� that can dissolve in water, highlighting its low solubility.
Dissolution Reaction
The process of barium sulfate dissolving in water can be expressed through a chemical reaction. This is often noted as the dissolution reaction. When BaSO鈧� dissolves, it dissociates into its constituent ions in water:\[\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)\]This equation tells us that for every molecule of solid BaSO鈧�, one barium ion and one sulfate ion are released into the solution. This dissolution reaches an equilibrium due to the low solubility. Hence, only a small concentration of ions remains in solution, aligning with the minimal solubility derived from calculations.
Stoichiometry
Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In the context of BaSO鈧� dissolving, stoichiometry helps us understand the 1:1 ratio in which the molecule dissociates into barium and sulfate ions. This means that any change in the concentration of BaSO鈧� directly influences the concentration of Ba虏鈦� and SO鈧劼测伝 ions in the solution.
  • 1 mole of BaSO鈧� produces 1 mole of Ba虏鈦� ions
  • 1 mole of BaSO鈧� produces 1 mole of SO鈧劼测伝 ions
When defining the solubility product constant, Ksp , stoichiometry is crucial. It helps us craft the expression Ksp = [Ba虏鈦篯[SO鈧劼测伝] , given their equal concentrations in the dissolution reaction.
Molarity
Molarity refers to the concentration of a solute in a solution, measured in moles per liter (M). Calculating molarity is vital to understand how much BaSO鈧� dissolves in water. In the given exercise:
  • The solubility translates to a molarity of \(1.07 \times 10^{-5} \text{ mol/L}\) for both Ba虏鈦� and SO鈧劼测伝 ions.
Knowing the molarity is essential when calculating the solubility product, Ksp, because it provides the concentrations of ions in solution. Having this data helps confirm the low solubility of BaSO鈧�, which aligns with its specific applications requiring low dissolution rates to minimize toxicity while delivering efficacy.

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Most popular questions from this chapter

If \(K_{\mathrm{c}}\) is greater than 1 for the reaction \(\mathrm{A} \rightleftarrows \mathrm{B}\), will the equilibrium concentrations of the products be smaller or larger than the equilibrium concentrations of the reactants? What if \(K_{\mathrm{c}}\) is less than \(1 ?\)

Use the equilibrium constants for reactions (a) and (b) at \(1000 \mathrm{~K}\) to calculate the equilibrium constant for reaction (c), the water-gas shift reaction, at that temperature. $$ \text { (a) } \begin{array}{c} \mathrm{CO}(g)+1 / 2 \mathrm{O}_{2}(g) \rightleftarrows \mathrm{CO}_{2}(g) \\ K_{\mathrm{c}}=1.1 \times 10^{18} \end{array} $$ (b) \(\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{H}_{2}(g)+1 / 2 \mathrm{O}_{2}(g)\) $$ \begin{array}{c} K_{\mathrm{c}}=7.1 \times 10^{-12} \\ \text {(c) } \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \\ K_{\mathrm{c}}=? \end{array} $$

Is the following statement true or false? The equilibrium concentrations depend on the initial concentrations, but the ratio of the equilibrium concentrations specified by the equilibrium constant expression is independent of the initial concentrations.

Write a chemical equation that describes the dissolution of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\). Write a mathematical equation that describes the relationship between the concentrations of the \(\mathrm{Ag}^{+}\) and \(\mathrm{CO}_{3}^{2-}\) ions in a saturated solution of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\).

Define the term common-ion effect. Describe how Le Ch芒telier's principle can be used to explain the common-ion effect.
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