Solutions

Define the basic terms and concepts related to preparing solutions of various concentrations, including part-to-part, percent, molarity, normality, and specific gravity.
Show an understanding of the most commonly used measures of concentration and why each is used where it is.
Calculate the proper quantity of a substance required to make a solution of a specified concentration, from pure stock and from previously diluted stock solutions.

Importance of accurate and precise chemical analyses in a laboratory setting.
Reagents or solutions can be prepackaged or need to be prepared from chemicals of recommended purity.
A solution is defined as a mixture of two or more substances that are not chemically combined.
Solute: Substance that is dissolved in a solution.
Solvent: Substance that dissolves the solute.
Solutions can be gaseous, liquid, or solid; focus is on liquid solutions in this chapter.
Concentration: Relative measurement of the amount of a substance in a solution, expressed in units such as grams per liter (g/L) and milligrams per deciliter (mg/dL).
Common expressions of concentration in the laboratory:
- Parts: Relationship of substances in a solution without specific units (e.g., 1 part A to 2 parts B).
- Percent: Defined as parts of solute per 100 parts of solution.
- Molarity (M): Concentration expressed as moles of solute in 1 L of solution.
  - Molarity formula: M = rac{ ext{moles of solute}}{1 ext{ L of solution}}
- Normality (N): Concentration of equivalents of solute in 1 L of solution.
  - Normality formula: N = rac{ ext{equivalents of solute}}{1 ext{ L of solution}}
- Specific Gravity: Measurement of density; relates mass of a substance to volume of an equal volume of water.

Parts allow for mixing based on volume or weight without specific units.
Example: Mixing one part chemical with three parts water could mean mixing 1 mL of bleach with 3 mL of water or any proportional volume/amount.
Importance of using the same units of measure throughout the preparation.
Example 1: Instructions for mixing herbicide with water.
- One part chemical to sixteen parts water.
- Calculation: Convert gallons to ounces (1 gal = 128 oz).
- Set up the equation: 16x = 128 to find x.
- Answer: 8 oz. of concentrated chemical mixed with 1 gal of water yields the correct solution strength.

A concentration expressed as a percentage is a ratio of solute parts to 100 parts of total solution.
Percent definitions:
- 5% w/v: rac{5 ext{ g}}{100 ext{ mL}}
- 5% v/v: rac{5 ext{ mL}}{100 ext{ mL}}
- 5% w/w: rac{5 ext{ g}}{100 ext{ g}}
Can use parts ratio to calculate amounts needed for specific volumes.
Example 1: Making a 10% w/v NaCl solution using 30 g of NaCl.
- Set up a proportion to find the total volume.
- Answer: 30 g of NaCl will make 300 mL of a 10% w/v NaCl solution.
Example 2: Preparation of a 15% w/w NaCl solution.
- Calculate the grams of NaCl needed for 500 g of solution.
- Answer: 75 g of NaCl and 425 g of H2O to produce the solution.

Definition of a mole: A measurement equivalent to 6.022 imes 10^{23} items (Avogadro's number).
A mole of a substance contains as many formula units as atoms in 12 g of carbon-12.
Molar Mass: Mass of 1 mol of atoms in grams.
Molarity (M): Concentration of a solution defined as moles of solute/L of solution.
To prepare a molar solution, use:
- Formula: ( ext{Gram Molecular Weight of Solute}) imes ( ext{Molarity Desired}) imes ( ext{Volume Needed}) = Grams of Solute Required.
- Conversion from milliliters to liters required for calculations.
Example 1: Making 500 mL of a 0.5 M saline solution requires measuring 14.61 g of NaCl.
To calculate molarity:
- Formula: M = rac{ ext{grams of solute}}{ ext{number of liters}} imes rac{1 ext{ mol}}{ ext{gmw}}
- Allows calculation of molarity from grams in a known volume.
Example 2: Calculating molarity of 150 g of salt in 500 mL using the same method resulting in an approximate concentration of 5.13 M.

Normality expresses concentration in equivalents of solute per liter of solution (eq/L).
Equivalent weight: Mass that will combine with/replaces 1 mol of hydrogen.
Normality is crucial for reactions, particularly acid-base reactions, where synaptic coupling takes place.
One equivalent of an acid = one equivalent of base in neutralization reactions.
Example 1: To prepare a 2 N solution of HCl in 1500 mL requires 109.35 g of HCl.

Specific gravity is the ratio of mass of a substance to the mass of an equal volume of water.
Example: H2SO4 with a specific gravity of 1.84 has an assay number of 97%, indicating it is primarily pure.
Specific gravity can convert grams to volume, making it easier for liquid solutions.

New solutions are created by diluting existing stock solutions.
Dilution is calculated with the formula: V₁C₁ = V₂C₂
Example: Producing a weaker solution from a stronger stock concentration.
Example 1: To produce 45 mL of 5% saline from 25 mL of 9%.
Cannot dilute a weaker solution to increase concentration—it's conceptually invalid.