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For example, sulfuric acid (H 2 SO 4) is a diprotic acid. Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be called a "2 normal" solution.
The solution has 1 mole or 1 equiv Na +, 1 mole or 2 equiv Ca 2+, and 3 mole or 3 equiv Cl −. An earlier definition, used especially for chemical elements , holds that an equivalent is the amount of a substance that will react with 1 g (0.035 oz) of hydrogen , 8 g (0.28 oz) of oxygen , or 35.5 g (1.25 oz) of chlorine —or that will displace ...
Cations are plotted in milliequivalents per liter on the left side of the zero axis, one to each horizontal axis, and anions are plotted on the right side. Stiff patterns are useful in making a rapid visual comparison between water from different sources. An alternative to the Stiff diagram is the Maucha diagram.
V eq is the volume of titrant (ml) consumed by the crude oil sample and 1 ml of spiking solution at the equivalent point, b eq is the volume of titrant (ml) consumed by 1 ml of spiking solution at the equivalent point, 56.1 g/mol is the molecular weight of KOH, W oil is the mass of the sample in grams. The normality (N) of titrant is calculated as:
Alkalinity is typically reported as mg/L as CaCO 3. (The conjunction "as" is appropriate in this case because the alkalinity results from a mixture of ions but is reported "as if" all of this is due to CaCO 3.) This can be converted into milliequivalents per Liter (meq/L) by dividing by 50 (the approximate MW of CaCO 3 divided by 2).
Where HV is the hydroxyl value; V B is the amount (ml) potassium hydroxide solution required for the titration of the blank; V acet is the amount (ml) of potassium hydroxide solution required for the titration of the acetylated sample; W acet is the weight of the sample (in grams) used for acetylation; N is the normality of the titrant; 56.1 is ...
For example, to make 100 mL of 50% alc/vol ethanol solution, water would be added to 50 mL of ethanol to make up exactly 100 mL. Whereas to make a 50% v/v ethanol solution, 50 mL of ethanol and 50 mL of water could be mixed but the resulting volume of solution will measure less than 100 mL due to the change of volume on mixing, and will contain ...
The peroxide value is defined as the amount of peroxide oxygen per 1 kilogram of fat or oil. Traditionally this was expressed in units of milliequivalents, although in SI units the appropriate option would be in millimoles per kilogram (N.B. 1 milliequivalents = 0.5 millimole; because 1 mEq of O 2 =1 mmol/2 of O 2 =0.5 mmol of O 2, where 2 is valence).