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Chapter 17: Problem 116

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The solubility product of strontium sulfate at 25°C is \(1.23 \times 10^{-6}\).

Step by step solution

01

Write the formula for osmotic pressure and the relationship between osmotic pressure and molar concentration.

The formula for osmotic pressure is: \(π = MRT\) where: - π: osmotic pressure (in this case, given as 21 torr) - M: molar concentration (in mol/L) - R: ideal gas constant (0.0821 L.atm/mol.K) - T: temperature in Kelvin (25°C = 298.15 K) In this case, we'll consider the osmotic pressure in atmospheres, so we'll need to convert 21 torr to atmospheres: \(1 atm = 760 torr \) so, \(π = \frac{21 torr}{760 torr/atm} = 0.02763 atm\)
02

Determine the molar concentration.

Using the osmotic pressure formula, we can now determine the molar concentration of strontium sulfate: \(M = \frac{π}{RT}\) Plugging in the values, we get: M = \(\frac{0.02763 atm}{(0.0821 L.atm/mol.K)(298.15 K)} = 1.11 \times 10^{-3} mol/L\)
03

Write the dissolution equation and solubility product expression for strontium sulfate.

The dissolution equation of strontium sulfate (\(SrSO_4\)) is: \(SrSO_{4(s)} \rightleftharpoons Sr^{2+}_{(aq)} + SO^{2-}_{4(aq)}\) The solubility product expression for strontium sulfate is: \(K_{sp} = [Sr^{2+}][SO^{2-}_{4}]\) Since the stoichiometric ratio is 1:1, the concentration of each ion in the solution is the same as the molar concentration of strontium sulfate. Thus, we have: \([Sr^{2+}] = [SO^{2-}_{4}] = 1.11 \times 10^{-3} mol/L\)
04

Calculate the solubility product of strontium sulfate at 25°C.

We can now plug in the concentrations into the solubility product expression to calculate the solubility product at 25°C: \(K_{sp} = (1.11 \times 10^{-3})(1.11 \times 10^{-3})\) \(K_{sp} = 1.23 \times 10^{-6}\) So, the solubility product of strontium sulfate at 25°C is \(\boxed{1.23 \times 10^{-6}}\).

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Key Concepts

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

Osmotic Pressure
Osmotic pressure is a key concept when studying solutions and their properties. It's the pressure required to prevent the flow of a solvent into a solution through a semipermeable membrane, which occurs due to osmosis—the movement of solvent molecules from a region of lower solute concentration to one of higher solute concentration.

The formula given in the exercise, \(\pi = MRT\), illustrates the direct proportionality between osmotic pressure (\(\pi\)), molar concentration (\(M\)) of the solute, the gas constant (\(R\)), and the absolute temperature (\(T\)).

Understanding this concept is crucial for grasping how solubility product relates to osmotic pressure. In the exercise, the conversion of osmotic pressure from torr to atmospheres was necessary to use standard units, which is important for the consistency and accuracy of calculations in chemistry.
Molar Concentration
The quantity known as molar concentration, denoted by \(M\), is a measure of the amount of a solute present in a given volume of solution, typically expressed in moles per liter (mol/L). It is a fundamental concept for understanding the concentration of particles in a solution and plays a vital role in chemical reactions and solution dynamics.

In the provided exercise, the molar concentration of strontium sulfate was calculated using the osmotic pressure formula where molar concentration can be rearranged to be \(M = \frac{\pi}{RT}\). This calculation shows the inverse relationship between molar concentration and both the temperature and the ideal gas constant. Comprehending how to find molar concentration is essential for students aiming to determine the solubility product, as it directly affects the values used in the dissolution equation.
Dissolution Equation
The dissolution equation represents the process by which a solid solute dissolves in a solvent to form a solution. It involves the breaking down of a substance into its constituent ions or molecules when it is dissolved in a solvent.

In the exercise, the dissolution equation for strontium sulfate (\(SrSO_4\)) is depicted by the equilibrium reaction: \(SrSO_{4(s)} \rightleftharpoons Sr^{2+}_{(aq)} + SO^{2-}_{4(aq)}\). This equation is necessary to understanding how to further calculate the solubility product (\(K_{sp}\)). The solubility product itself is an equilibrium constant that applies to the saturated solution of an ionic compound, as seen with strontium sulfate.

Knowing each ion's concentration in the solution allows us to calculate \(K_{sp}\), by simply multiplying the concentrations of the ions as shown in the solution step: \(K_{sp} = [Sr^{2+}][SO^{2-}_{4}]\). The dissolution equation is imperative in predicting the solubility of compounds and understanding the dynamics of ionic substances in solution.

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Most popular questions from this chapter

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\)) reacts with acids in foods to form carbonic acid (\(\mathrm{H}_{2} \mathrm{CO}_{3}\)), which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(g)\) forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that 1\(/ 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH (\mathrm{OH}) \)\mathrm{COOH}\( .Write the chemical equation for this neutralization reaction. (b) The density of baking soda is 2.16 \)\mathrm{g} / \mathrm{cm}^{3} .\( Calculate the concentration of lactic acid in one cup of sour milk(assuming the rule of thumb applies), in units of mol/L. (One cup \)=236.6 \mathrm{mL}=48\( teaspoons). (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to \)350^{\circ} \mathrm{F}$ .

Consider a beaker containing a saturated solution of \(\mathrm{CaF}_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s)\). Solid \(\mathrm{CaCl}_{2}\) is then added to the solution. (a) Will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(\mathbf{F}^{-}\) ions in solution increase or decrease?

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 -mL sample titrated with HCl requires 11.23 \(\mathrm{mL}\) of 0.0983\(M \mathrm{HCl}\) to reach the end point. Calculate \(K_{s p}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\) . Compare your result with that in Appendix D. Suggest a reason for any differences you find between your value and the one in Appendix D.

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{A}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a 1.0 \(\mathrm{M}\) solution of this acid and a 1.0 \(\mathrm{M}\) solution of \(\mathrm{NaOH} .\) How much of the NaOH solution should be added to 1.0 \(\mathrm{L}\) of the acid to give a buffer at pH 6.50\(?\) (Ignore any volume change.)

Consider a beaker containing a saturated solution of \(\mathrm{PbI}_{2}\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s)\). Now solid \(\mathrm{K} \mathrm{I}\) is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(I\) ions in solution increase or decrease?
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