What is the osmotic pressure of a \(0.250 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) at \(25^{\circ} \mathrm{C}\) ? (Assume complete dissociation.)

Short Answer

Expert verified
Answer: The osmotic pressure of a 0.250 M solution of K2CO3 at 25 °C is approximately 55.00 atm.

Step by step solution

01

Determine the number of particles after dissociation

K2CO3 is a salt that will completely dissociate into its constituent ions. In this case, one K2CO3 molecule will dissociate into two potassium ions (K⁺) and one carbonate ion (CO₃²⁻). Therefore, for every 1 mole of K2CO3, there will be a total of 3 moles of particles in the solution.
02

Calculate the total concentration of particles

We are given that the initial concentration of K2CO3 is 0.250 M. Since it dissociates into a total of 3 moles of particles per mole of K2CO3, we can multiply that initial concentration by the number of particles formed: Total concentration of particles = 0.250 M × 3 = 0.750 M
03

Use van't Hoff's equation to find osmotic pressure

Van't Hoff's equation for osmotic pressure is given by: π=iMRT where π = osmotic pressure, i = van't Hoff factor (number of particles in the solution, which is 3 in this case), M = molar concentration (calculated in step 2 as 0.750 M), R = ideal gas constant (0.0821 L atm / K mol), T = temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin: T(K) = 25 °C + 273.15 = 298.15 K Now, we can use all these values to find the osmotic pressure: π = (3)(0.750 M)(0.0821 L atm / K mol)(298.15 K)
04

Calculate the osmotic pressure

Calculating the osmotic pressure using the values obtained in the previous steps: π = (3)(0.750 M)(0.0821 L atm / K mol)(298.15 K) ≈ 55.00 atm Thus, the osmotic pressure of a 0.250 M solution of K2CO3 at 25 °C is approximately 55.00 atm.

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Key Concepts

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

van't Hoff's equation
Understanding the concept of osmotic pressure can be facilitated by a critical look at van't Hoff's equation. In basic terms, van't Hoff's equation is the formula used to calculate the osmotic pressure exerted by solutes in a solution. It is represented as \( \pi = iMRT \), where \( \pi \) stands for osmotic pressure, \( i \) is the van't Hoff factor indicating the number of particles into which a solute dissociates, \( M \) is the molar concentration of the solute, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

The van't Hoff factor, \( i \) is crucial because it accounts for the actual number of particles that a compound breaks into when dissolved. As a rule of thumb, for most non-electrolytes (compounds that do not dissociate into ions), \( i \) is taken as 1. However, for electrolytes like \( \mathrm{K}_{2} \mathrm{CO}_{3} \) used in our exercise, \( i \) is not 1 but the total number of ions produced upon dissociation - which affects the calculated osmotic pressure.
Dissociation of ionic compounds
A key factor in calculating osmotic pressure is the understanding of dissociation in ionic compounds. Ionic compounds are made up of charged ions held together by ionic bonds. When these compounds are dissolved in a solvent such as water, they dissociate into their individual ions. For instance, \( \mathrm{K}_{2} \mathrm{CO}_{3} \) dissociates into two \( \mathrm{K}^{+} \) ions and one \( \mathrm{CO}_{3}^{2-} \) ion.

Dissociation significantly influences the number of particles in a solution, which is directly proportional to the osmotic pressure. If dissociation is complete, as assumed in the problem provided, the total number of particles affects the van't Hoff factor \( i \). Thus, accurately identifying the number of particles resulting from dissociation is a critical step in calculating the correct osmotic pressure.
Molar concentration
Molar concentration, often represented by the symbol \( M \), measures the amount of substance per unit volume of solution, and it's expressed in moles per liter (mol/L). It is an essential parameter in the van't Hoff equation for osmotic pressure. To obtain the total molar concentration of particles in a solution after dissociation, you must multiply the molar concentration of the ionic compound by the number of particles formed after dissociation.

In the given exercise, starting with a 0.250 M solution of \( \mathrm{K}_{2} \mathrm{CO}_{3} \) means you initially have 0.250 moles of \( \mathrm{K}_{2} \mathrm{CO}_{3} \) in every liter of solution. After dissociation into three ions, the total molar concentration of particles becomes 0.750 M. This adjusted concentration is then used to determine the osmotic pressure, revealing the colligative property of the solution, which depends on the number of solute particles rather than their identity.
Ideal gas constant
The ideal gas constant, denoted as \( R \), is a physical constant that appears in many fundamental equations in the physical sciences, such as the ideal gas law and the van't Hoff equation. Its value is determined empirically and is the same for all gases, assuming ideal behavior. The constant provides a link between the macroscopic and microscopic properties of gases.

When used in the context of calculating osmotic pressure via van't Hoff's equation, \( R \) provides the proportionality between the number of moles, temperature, and pressure. Its value is given as 0.0821 atm L/mol K, which ensures that the pressure is calculated in atmospheres when the remaining units of concentration and temperature are in moles per liter and Kelvin, respectively. This constant is essential and ensures the various units within the equation are consistent, allowing for the correct calculation of osmotic pressure.

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Most popular questions from this chapter

A certain gaseous solute dissolves in water, evolving \(12.0 \mathrm{~kJ}\) of heat. Its solubility at \(25^{\circ} \mathrm{C}\) and \(4.00\) atm is \(0.0200 M .\) Would you expect the solubility to be greater or less than \(0.0200 M\) at (a) \(5^{\circ} \mathrm{C}\) and 6 atm? (b) \(50^{\circ} \mathrm{C}\) and 2 atm? (c) \(20^{\circ} \mathrm{C}\) and 4 atm? (d) \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} ?\)

A bottle of commercial concentrated aqueous ammonia is labeled "29.89\% \(\mathrm{NH}_{3}\) by mass; density \(=0.8960 \mathrm{~g} / \mathrm{mL}\)." (a) What is the molarity of the ammonia solution? (b) If \(250.0 \mathrm{~mL}\) of the commercial ammonia is diluted with water to make \(3.00 \mathrm{~L}\) of solution, what is the molarity of the diluted solution?

A sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) solution that is \(45.0 \%\) sucrose by mass has a density of \(1.203 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate its (a) molarity. (b) molality. (c) vapor pressure \(\left(\mathrm{vp} \mathrm{H}_{2} \mathrm{O}\right.\) at \(25^{\circ} \mathrm{C}=23.76 \mathrm{~mm} \mathrm{Hg}\) ). (d) normal boiling point.

A carbonated beverage is made by saturating water with carbon dioxide at \(0^{\circ} \mathrm{C}\) and a pressure of \(3.0 \mathrm{~atm}\). The bottle is then opened at room temperature \(\left(25^{\circ} \mathrm{C}\right)\), and comes to equilibrium with air in the room containing \(\mathrm{CO}_{2}\left(P_{\mathrm{CO}_{2}}=3.4 \times 10^{-4} \mathrm{~atm}\right)\). The Henry's law constant for the solubility of \(\mathrm{CO}_{2}\) in water is \(0.0769\) M/atm at \(0^{\circ} \mathrm{C}\) and \(0.0313 \mathrm{M} / \mathrm{atm}\) at \(25^{\circ} \mathrm{C}\). (a) What is the concentration of carbon dioxide in the bottle before it is opened? (b) What is the concentration of carbon dioxide in the bottle after it has been opened and come to equilibrium with the air?

The Henry's law constant for the solubility of oxygen in water is \(3.30 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) at \(12^{\circ} \mathrm{C}\) and \(2.85 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) at \(22^{\circ} \mathrm{C}\). Air is \(21 \mathrm{~mol} \%\) oxygen. (a) How many grams of axygen can be dissolved in one liter of a trout stream at \(12^{\circ} \mathrm{C}\left(54^{\circ} \mathrm{F}\right)\) at an air pressure of \(1.00 \mathrm{~atm} ?\) (b) How many grams of oxygen can be dissolved per liter in the same trout stream at \(22^{\circ} \mathrm{C}\left(72^{\circ} \mathrm{F}\right)\) at the same pressure as in (a)? (c) A nuclear power plant is responsible for the stream's increase in temperature. What percentage of dissolved oxygen is lost by this increase in the stream's temperature?

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