Concept of solution in chemistry

A solution is a homogeneous mixture of two or more substances, where one substance (the solute) is dissolved in another substance (the solvent). Solutions are essential in various scientific fields and industries, particularly in chemistry, biology, and pharmaceuticals. Understanding the preparation and concentration of solutions is crucial for accurate experimentation and product development. Below are key concepts related to solutions:

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Key Characteristics of Solutions

Homogeneity:

• A solution is a homogeneous mixture, meaning that the solute is uniformly distributed within the solvent. You cannot see the individual particles of solute with the naked eye.

Solvent and Solute:

• Solvent: The substance that dissolves the solute. In most cases, water is the solvent (forming aqueous solutions), but it can also be other liquids or gases.
• Solute: The substance that gets dissolved. It can be a solid, liquid, or gas.

Phase:

• Solutions can be solid, liquid, or gas, depending on the phases of the solute and solvent. For example, in an aqueous solution, the solvent is water (liquid), and the solute might be a solid (like salt) or gas (like oxygen).

Molecular Level Dispersion:

• In a solution, the solute particles are broken down into individual molecules, atoms, or ions and dispersed evenly within the solvent. This prevents the solute from forming separate phases or layers.

Stability:

• A solution is stable, meaning that the solute does not settle out over time, unlike in suspensions or colloids. The solute remains uniformly distributed as long as the conditions remain unchanged (temperature, pressure).

Types of Solutions

Solutions can be categorized based on their physical states, the type of solute and solvent, or their concentration levels. Understanding these classifications helps in identifying and analyzing the behavior of different solutions in various chemical and practical applications.

1. Based on Physical State

(a) Solid Solutions

• Solute: Solid, Liquid, or Gas
• Solvent: Solid
• Examples:
  • Alloys like brass (zinc in copper).
  • Hydrogen gas dissolved in palladium.

(b) Liquid Solutions

• Solute: Solid, Liquid, or Gas
• Solvent: Liquid
• Examples:
  • Sugar dissolved in water (solid in liquid).
  • Alcohol in water (liquid in liquid).
  • Carbon dioxide in soda (gas in liquid).

(c) Gaseous Solutions

• Solute: Gas, Liquid, or Solid
• Solvent: Gas
• Examples:
  • Air (oxygen and other gases dissolved in nitrogen).
  • Humid air (water vapor in air).

2. Based on Solvent Type

(a) Aqueous Solutions

• Solutions where water acts as the solvent.
• Examples:
  • Saltwater.
  • Vinegar (acetic acid in water).

(b) Non-Aqueous Solutions

• Solutions where the solvent is not water.
• Examples:
  • Iodine dissolved in alcohol (tincture of iodine).
  • Sulfur in carbon disulfide.

3. Based on Solute-Solvent Interaction

(a) Electrolytic Solutions

• Contain ionic solutes that dissociate into ions and conduct electricity.
• Examples:
  • Sodium chloride in water.
  • Acids like HCl dissolved in water.

(b) Non-Electrolytic Solutions

• Contain covalent solutes that do not produce ions and do not conduct electricity.
• Examples:
  • Sugar in water.
  • Ethanol in water.

4. Based on Concentration

(a) Dilute Solutions

• Contain a small amount of solute relative to the solvent.
• Example: 0.1 M NaCl solution.

(b) Concentrated Solutions

• Contain a large amount of solute relative to the solvent.
• Example: Saturated salt solution.

(c) Saturated Solutions

• Contain the maximum amount of solute that can dissolve at a given temperature and pressure.
• Example: Saltwater at room temperature with undissolved salt at the bottom.

(d) Unsaturated Solutions

• Contain less solute than the saturation point, meaning more solute can still dissolve.
• Example: 5 g of sugar dissolved in 100 ml of water.

(e) Supersaturated Solutions

• Contain more solute than the saturation point, achieved by changing temperature or pressure.
• Example: Sodium acetate solution cooled slowly after saturation.

5. Based on Particle Size

(a) True Solutions

• Homogeneous mixtures where solute particles are molecular or ionic in size (<1 nm).
• Examples: Saltwater, sugar in water.

(b) Colloidal Solutions

• Heterogeneous mixtures with particle sizes between 1 nm and 100 nm.
• Examples: Milk, gelatin.

(c) Suspensions

• Heterogeneous mixtures with large solute particles (>100 nm) that settle out over time.
• Examples: Mud in water, flour in water.

6. Based on Nature of Solute

(a) Volatile Solutions

• Contain solutes that easily vaporize.
• Examples: Alcohol in water.

(b) Non-Volatile Solutions

• Contain solutes that do not vaporize easily.
• Examples: Salt in water.

7. Based on Miscibility (For Liquids)

(a) Miscible Solutions

• Liquids that dissolve completely in each other.
• Examples: Ethanol in water.

(b) Immiscible Solutions

• Liquids that do not dissolve in each other.
• Examples: Oil in water.

Properties of Solutions 

Solutions exhibit unique properties that distinguish them from pure substances and other types of mixtures. These properties can be categorized as colligative properties, physical properties, and chemical properties.

Colligative Properties

Colligative properties depend on the number of solute particles in a solution, not their chemical identity. These properties include:

1. Vapor Pressure Lowering:

• The presence of solute particles reduces the vapor pressure of the solvent.
• This occurs because solute particles disrupt the evaporation process at the surface of the liquid.

2. Boiling Point Elevation:

• The boiling point of a solution is higher than that of the pure solvent.
• This happens because additional energy is required to overcome the solute-solvent interactions and convert the liquid into vapor.

$$\mathrm{\Delta }{T}_b=i\cdot K_b\cdot m$$

where:

• $\mathrm{\Delta }{T}_b$: Boiling point elevation
• $i$: Van’t Hoff factor
• $K_b$: Molal boiling point elevation constant
• $m$: Molality of the solution

3. Freezing Point Depression:

• The freezing point of a solution is lower than that of the pure solvent.
• Solute particles interfere with the formation of the solid phase, requiring a lower temperature for freezing.

$$\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m$$

where:

• $\mathrm{\Delta }{T}_f$: Freezing point depression
• $K_f$: Molal freezing point depression constant

4. Osmotic Pressure:

• The pressure required to stop the flow of solvent through a semipermeable membrane from a dilute solution to a concentrated solution.
• Osmotic pressure ($\mathrm{\Pi }$) is calculated as:

$$\mathrm{\Pi }=i\cdot M\cdot R\cdot T$$

where:

• $M$: Molarity of the solution
• $R$: Ideal gas constant
• $T$: Temperature in Kelvin

Physical Properties

1. Homogeneity:

• Solutions are uniform throughout, meaning the solute particles are evenly distributed in the solvent.

2. Particle Size:

• The solute particles are typically molecular or ionic in size, making them too small to be seen with the naked eye or to settle out over time.

3. Transparency:

• Most solutions are transparent because the solute particles do not scatter light (exceptions include colloidal solutions).

4. Conductivity:

• Solutions containing electrolytes (like salts) conduct electricity due to the presence of ions, whereas non-electrolyte solutions (like sugar in water) do not.

Chemical Properties

1. Reactivity:

• The chemical behavior of a solution can depends on the nature of the solute and solvent. For instance, acids and bases exhibit characteristic reactions in solution.

2. Solubility:

• Solubility refers to the maximum amount of solute that can dissolve in a given quantity of solvent at a specific temperature and pressure.
• Factors affecting solubility include:
  • Nature of the solute and solvent (polar or non-polar).
  • Temperature: Solubility of solids typically increases with temperature, while gases often become less soluble.
  • Pressure: Primarily affects the solubility of gases (e.g., Henry’s Law).

3. Chemical Equilibria:

• Solutions can participate in chemical equilibria, such as acid-base equilibria or redox reactions, which are influenced by the concentration of solutes.

Preparation of Solution

The preparation of a solution involves dissolving a specific amount of solute in a solvent to create a homogeneous mixture. The concentration of the solution depends on the amount of solute dissolved in a given volume of solvent. This process typically involves the following steps:

1. Weighing the solute: Accurately measure the amount of solute using a balance.
2. Dissolving the solute: Add the solute to a suitable solvent and stir or shake to ensure it dissolves completely.
3. Adjusting the volume: If preparing a solution of a specific concentration, adjust the volume by adding more solvent until the desired volume is reached.

Dilute Solution

A dilute solution is one in which the concentration of the solute is relatively low compared to the solvent. Diluting a concentrated solution involves adding more solvent to reduce the concentration of solute. This can be done for various purposes, such as for safety reasons in a laboratory or to achieve a required concentration for an experiment.

To prepare a dilute solution, the following steps are generally followed:

1. Use of concentrated solution: Start with a more concentrated solution.
2. Calculation of dilution factor: Use the formula $C_1{V}_1=C_2{V}_2$, where $C_1$ and $V_1$ are the concentration and volume of the concentrated solution, and $C_2$ and $V_2$ are the concentration and volume of the diluted solution.
3. Add solvent: Slowly add solvent to the concentrated solution until the desired concentration is achieved.

Sample Questions and Answers on Dilution of Solution 

Question 1: 

To determine the new concentration of the potassium hydroxide solution after dilution, we use the concept of dilution, where the moles of solute remain constant before and after dilution. The relationship can be expressed as:

$$C_1{V}_1=C_2{V}_{\mathrm{2}}$$

Where:

• $C_1$ is the initial concentration (2 mol·dm⁻³),
• $V_1$ is the initial volume (25 cm³ or 0.025 dm³),
• $C_2$ is the final concentration (to be calculated),
• $V_2$ is the final volume (500 cm³ or 0.5 dm³).

Step 1: Solve for C2​

$$C_2 = \frac{C_1 V_1}{V_2}$$

Substitute the values:

$$C_2 = \frac{2 \, \text{mol·dm⁻³} \times 0.025 \, \text{dm³}}{0.5 \, \text{dm³}}$$

Step 2: Calculate${\mathrm{ C2}}_{}$

$$C_2 = \frac{0.05}{0.5} = 0.1 \, \text{mol·dm⁻³}$$

Final Answer:

The new concentration of the solution after dilution is 0.1 mol·dm⁻³.

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Question 2: 

What volume of $15 \, \text{mol·dm⁻³}$ sulfuric acid ($H_2SO_4$) solution should be diluted to $1 \, \text{dm³}$ to make a $0.1 \, \text{mol·dm⁻³}$ solution of the acid? Show your calculations.

Solution:

We use the dilution formula:

$$C_1{V}_1=C_2{V}_{\mathrm{2}}$$

Where:

• $C_1 = 15 \, \text{mol·dm⁻³}$ (initial concentration),
• $V_1$ is the volume of the concentrated solution to be calculated,
• $C_2 = 0.1 \, \text{mol·dm⁻³}$ (final concentration),
• $V_2 = 1 \, \text{dm³}$ (final volume).

Step 1: Solve for V1​

Rearranging the formula:

$$V_1=\frac{C_2{V}_2}{C_{\mathrm{1<span\; class="vlist"><span\; class="mord"><span\; class="mord"><span\; class="msupsub"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span></span></span></span><span\; class="vlist-s"></span>}}}$$

Substitute the known values:

$$V_1=\frac{0.1\text{moll.dm⁻³}\times 1\text{dm³}}{15\text{mol.}\text{dm⁻³}}$$

Step 2: Calculate V1​

$$V_1 = \frac{0.1}{15} = 0.00667 \, \text{dm³}$$

Convert to cm³ (1 dm³ = 1000 cm³):

$$V_1=0.00667\text{dm³}\times 1000=6.67\text{cm³}$$

Final Answer:

The volume of sulfuric acid needed is 6.67 cm³.

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Question 3: 

If $100 \, \text{cm}^3$ of $0.2 \, \text{mol·dm}^{-3}$ hydrochloric acid ($HC\mathrm{l)\; is\; diluted\; to }$$500 \, \text{cm}^3$, calculate:

1. The dilution factor.
2. The concentration of the new solution in $\text{mol·dm}^{-3}$.

Solution:

(i) Dilution Factor

The dilution factor is the ratio of the final volume to the initial volume. It is calculated as:

$$\text{Dilution Factor}=\frac{\text{Final Volume}}{\text{Initial Volume}}$$

Given:

• Final Volume = $500 \, \text{cm}^3$,
• Initial Volume = $100 \, \text{cm}^3$,

$$\text{Dilution Factor} = \frac{500}{100} = 5$$

(ii) Concentration of the New Solution

Using the dilution relationship:

$$C_1{V}_1=C_2{V}_{\mathrm{2}}$$

Where:

• $C_1 = 0.2 \, \text{mol·dm}^{-3}$ (initial concentration),
• $V_1 = 100 \, \text{cm}^3 = 0.1 \, \text{dm}^3$ (initial volume),
• $C_2$ is the final concentration (to be calculated),
• $V_2 = 500 \, \text{cm}^3 = 0.5 \, \text{dm}^3$ (final volume).

Rearranging for $C_2$:

$$C_2=\frac{C_1{V}_1}{V_{\mathrm{2<span\; class="vlist-s"></span>}}}$$

Substitute the known values:

$$C_2 = \frac{0.2 \, \text{mol·dm}^{-3} \times 0.1 \, \text{dm}^3}{0.5 \, \text{dm}^3}$$ $$C_2 = \frac{0.02}{0.5} = 0.04 \, \text{mol·dm}^{-3}$$

Final Answer:

1. The dilution factor is 5.
2. The concentration of the new solution is 0.04 mol·dm⁻³.

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Question 4: 

How much distilled water should be added to $400 \, \text{cm}^3$ of $2.50 \, \text{mol·dm}^{-3}$ sulfuric acid ($H_2SO_4$) to obtain a $0.250 \, \text{mol·dm}^{-3}$ solution?

Solution:

We use the dilution formula:

$$C_1{V}_1=C_2{V}_{\mathrm{2}}$$

Where:

• $C_1 = 2.50 \, \text{mol·dm}^{-3}$ (initial concentration),
• $V_1 = 400 \, \text{cm}^3 = 0.4 \, \text{dm}^3$ (initial volume),
• $C_2 = 0.250 \, \text{mol·dm}^{-3}$ (final concentration),
• $V_2$ is the final volume (to be calculated).

Step 1: Solve for V2​

Rearranging the formula for $V_2$:

$$V_2=\frac{C_1{V}_1}{C_{\mathrm{2<span\; class="vlist-s"></span>}}}$$

Substitute the known values:

$$V_2 = \frac{2.50 \times 0.4}{0.250}$$ $$V_2 = \frac{1.0}{0.250} = 4.0 \, \text{dm}^3$$

Step 2: Calculate the Volume of Water to Add

The total final volume is $4.0 \, \text{dm}^3 = 4000 \, \text{cm}^3$. The initial volume is $400 \, \text{cm}^3$. The volume of distilled water to add is:

$$\text{Volume of Water} = V_2 - V_1$$ $$\text{Volume of Water} = 4000 \, \text{cm}^3 - 400 \, \text{cm}^3 = 3600 \, \text{cm}^3$$

Final Answer:

You need to add 3600 cm³ of distilled water to the solution.

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Question 5: 

100 cm³ of distilled water was added to 400 cm³ of 0.5 mol·dm⁻³ hydrochloric acid (HCl) solution and stirred thoroughly. What is the final concentration in:

1. Moles per dm³
2. Grams per dm³ (Given: H = 1, Cl = 35.5)

Solution:

Given:

• Initial volume of HCl solution, $V_1 = 400 \, \text{cm}^3 = 0.4 \, \text{dm}^3$,
• Initial concentration of HCl, $C_1 = 0.5 \, \text{mol·dm}^{-3}$,
• Volume of distilled water added = $100 \, \text{cm}^3$,
• Final volume $V_2 = 400 \, \text{cm}^3 + 100 \, \text{cm}^3 = 500 \, \text{cm}^3 = 0.5 \, \text{dm}^3$.

Step 1: Final Concentration in Moles per dm³

We use the dilution formula:

$$C_1{V}_1=C_2{V}_2$$

Where:

• $C_1 = 0.5 \, \text{mol·dm}^{-3}$ (initial concentration),
• $V_1 = 0.4 \, \text{dm}^3$ (initial volume),
• $C_2$ is the final concentration (to be calculated),
• $V_2 = 0.5 \, \text{dm}^3$ (final volume).

Rearranging for $C_2$:

$$C_2=\frac{C_1{V}_1}{V_{\mathrm{2<span\; class="vlist-s"></span>}}}$$

Substitute the values:

$$C_2 = \frac{0.5 \, \text{mol·dm}^{-3} \times 0.4 \, \text{dm}^3}{0.5 \, \text{dm}^3}$$ $$C_2=\frac{0.2}{0.5}=0.4{\text{mol.}\text{dm}}^{-3}$$

Thus, the final concentration in moles per dm³ is 0.4 mol·dm⁻³.

Step 2: Final Concentration in Grams per dm³

First, we need to calculate the molar mass of HCl:

$$\text{Molar mass of HCl}=1+35.5=36.5\text{g/mol}$$

Now, using the formula:

$$\text{Concentration in grams per dm³}=C_2\times \text{Molar mass of HCl}$$

Substitute the values:

$$\text{Concentration in grams per dm³} = 0.4 \, \text{mol·dm}^{-3} \times 36.5 \, \text{g/mol}$$
$$\text{Concentration in grams per dm³}=14.6{\text{g.}\text{dm}}^{-3}$$

Thus, the final concentration in grams per dm³ is 14.6 g·dm⁻³.

Final Answer:

1. Moles per dm³: 0.4 mol·dm⁻³
2. Grams per dm³: 14.6 g·dm⁻³

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Question 6: 

A 250 cm³ solution of $K_2Cr_2O_7$ with concentration $0.025 \, \text{mol·dm}^{-3}$ is to be prepared from a stock solution with concentration $0.04 \, \text{mol·dm}^{-3}$. Calculate the volume of the stock solution required.

Solution:

We use the dilution formula:

$$C_1{V}_1=C_2{V}_{\mathrm{2}}$$

Where:

• $C_1 = 0.04 \, \text{mol·dm}^{-3}$ (stock solution concentration),
• $V_1$ is the volume of the stock solution to be calculated,
• $C_2 = 0.025 \, \text{mol·dm}^{-3}$ (final solution concentration),
• $V_2 = 250 \, \text{cm}^3 = 0.25 \, \text{dm}^3$(final volume).

Step 1: Solve for V1​

Rearrange the formula for $V_1$:

$$V_1=\frac{C_2{V}_2}{C_{\mathrm{1<span\; class="vlist-s"></span>}}}$$

Substitute the known values:

$$V_1=\frac{0.025{\text{mol.}\text{dm}}^{-3}\times 0.25{\text{dm}}^3}{0.04{\text{mol.}\text{dm}}^{-\mathrm{3}}}$$

Step 2: Calculate V1​

$$V_1 = \frac{0.00625}{0.04} = 0.15625 \, \text{dm}^3$$

Convert to cm³ (1 dm³ = 1000 cm³):

$$V_1 = 0.15625 \, \text{dm}^3 \times 1000 = 156.25 \, \text{cm}^3$$

Final Answer:

The volume of the stock solution required is 156.25 cm³.

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