Introduction to Chemical Concepts
- MOLARITY (M):
- by definition, 1 mole per liter.
- Solvent is added and brought up to 1 L
- MOLALITY (m):
- by definition, 1 mole per kg of solvent
- for example, an aqueous solution with water, 1 kg of H20 is approximately 1 L
- 1mole is added to 1 kg of solvent
- NORMALITY (N):
- by definition, one equivalent per liter
- What is an equivalent?
- Molecular weight divided by the number of replaceable protons. For example, HCl there is one replaceable proton and therefore normality = molarity
- In contrast, H2SO4 or H3PO4 have two and three replaceable protons, respectively. In the case of H2SO4, a 1 N solution would be equal to a 0.5 M solution.
- The advantage of normality is in titration, 1 equivalent of acid will completely normalize one equivalent of base.
- EQUILAVENTS (Eq):
- Electrolyte concentrations are expressed in milliequivalents (mEq)
- For single elements, an equivalent is the molecular weight divided by the absolute value of the oxidative state.
- Na+1: 23 gm = 1 Eq and 23 mg = 1 mEq
- Cl-1: 35.5 gm = 1 Eq and 35.5 mg= 1 mEq
- Ca+2: 20.0 gm = 1 Eq and 20 mg = 1 mEq
- For compounds, equivalents are calculated as the molecular weight divided by the absolute value of the highest oxidative state
- NaCl: 58.5 gm = 1 Eq NaCl
- CaCl2: 55.5 gm = 1 Eq CaCl2
- AlCl3: 44 gm = 1 Eq AlCl2
- Equivalent weights do not always remain the same since they result from the net electron change
- PERCENT COMPOSITION SOLUTIONS
- weight/weight (w/w)
- gm of solute per gm of solvent
- Example:
- 10% (w/w) NaOH: 10 gm of NaOH and 90 gm H20
- weight/volume (w/v)
- Example:
- 10% (w/v) NaOH: 10 gm of NaOH dissolved in a small amount of solvent then brought up to 100 mL
- Example:
- volume/volume (v/v)
- Example:
- 30% Ethanol: 30 mL of 100% Ethanol and 70 mL H20
- Example:
- weight/weight (w/w)
- CLINICAL EXAMPLES
- mg % = mg/dL = mg/100 mL
- TRACE METALS
- ppm (parts per million): mg / L
- ppb (parts per billion) µg / L