Solution Solute Solvent Molar Molarity Concentration Dilution

Overview

Engineering reference data for Solution Solute Solvent Molar Molarity Concentration Dilution in chemistry.

Key Formulas

Ideal Gas Law

$PV = nRT$

Pressure × Volume = moles × gas constant × temperature.

Molarity

$M = \frac{n}{V}$

Moles of solute per liter of solution.

pH

$pH = -\log_{10}[H^+]$

Measure of acidity.

Variables

Factors Affecting Solubility

Solubility is influenced by several factors:

• Temperature: In general, solubility increases with temperature, though there are exceptions.
• Particle Size: Smaller particles dissolve faster than larger particles.
• Agitation: Stirring or shaking a mixture increases the rate at which a solid dissolves.

Concentration Expressions

The concentration of a solute in a solution can be expressed in several ways:

Worked Examples

Example 1: Calculating Mass Percent

Problem: What is the concentration (in wt%) of a solution where 56 g of salt is dissolved in 0.8 kg of water?

Solution:

1. Convert water mass to grams: 0.8 kg = 800 g
2. Calculate total solution mass: 56 g + 800 g = 856 g
3. Calculate mass percent: $\text{wt\%} = \frac{56 \text{ g}}{856 \text{ g}} \times 100\% \approx 6.54\%$

Example 2: Saturated Solutions and Solubility Limit

Problem: Sodium chloride (NaCl) has a solubility of 36 g per 100 g of water at 25°C. Calculate the concentration for different amounts added.

Solution:

1. Saturated solution (36 g in 100 g water): $\text{wt\%} = \frac{36 \text{ g}}{36 \text{ g} + 100 \text{ g}} \times 100\% = 26.5\%$
2. 34 g added (below solubility limit): All salt dissolves. $\text{wt\%} = \frac{34 \text{ g}}{34 \text{ g} + 100 \text{ g}} \times 100\% = 25.4\%$
3. 38 g added (above solubility limit): Solution becomes saturated; excess remains undissolved. Concentration is the same as the saturated solution: 26.5 wt%.

Example 3: Molarity Calculation

Problem: Calculate the molarity of a solution made by dissolving 200 g of NaCl in enough water to make 5.0 L of solution. (Molar masses: Na = 22.99 g/mol, Cl = 35.45 g/mol)

Solution:

1. Calculate the molar mass of NaCl: $M_{\text{NaCl}} = 22.99 + 35.45 = 58.44 \text{ g/mol}$
2. Calculate moles of NaCl: $n = \frac{200 \text{ g}}{58.44 \text{ g/mol}} \approx 3.422 \text{ mol}$
3. Calculate molarity: $M = \frac{3.422 \text{ mol}}{5.0 \text{ L}} = 0.684 \text{ M}$

References

• Original Source