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Handling, Preparation and Storage of Standards

ICP Operations Guide:
Part 3

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Calculations

The concentration units for chemical standard solutions used for ICP applications are typically expressed in µg/mL (micrograms per milliliter) or ng/mL (nanograms per milliliter). For example, a 1000 µg/mL solution of Ca+2 contains 1000 micrograms of Ca+2 per each mL of solution and a 1 µg/mL solution of Ca+2 contains 1000 ng of Ca+2 per milliliter of solution. To convert between metric concentration units the following conversions apply:

Table 3.1: Mass portion of concentration unit where g = gram

 Prefix  Scientific Notation  Decimal equivalents  Example Units
 kilo- (k)   = 103   = 1000 g   kilogram (kg)
 milli- (m)   = 10-3   = 0.001 g   milligram (mg)
 micro- (µ)   = 10-6   = 0.000001 g   microgram (µg)
 nano- (n)   = 10-9   = 0.000000001 g   nanogram (ng)
 pico- (p)   = 10-12   = 0.000000000001 g   picogram (pg)

Table 3.2: Volume portion of concentration unit where L = liter

 Prefix  Scientific Notation  Decimal equivalents  Example Units
 milli- (m)   = 10-3   = 0.001 L   milliliter (mL)
 micro- (µ)   = 10-6   = 0.000001 L   microliter (µL)
 nano- (n)   = 10-9   = 0.000000001 L   nanoliter (nL)
 pico- (p)   = 10-12   = 0.000000000001 L   picoliter (pL)

The difference between ppm and µg/mL is often confused. A common mistake is to refer to the concentration units in ppm as a short cut (parts per million) when we really mean µg/mL. One ppm is in reality equal to 1 µg/g. In similar fashion ppb (parts per billion) is often equated with ng/mL. One ppb is in reality equal to 1 ng/g. To convert between ppm or ppb to µg/mL or ng/mL the density of the solution must be known. The equation for conversion between wt./wt. and wt./vol. units is:

(µg/g) (density in g/mL) = µg/mL

and/or

(ng/g) (density in g/mL) = ng/mL

Therefore, if we have a solution that is 1000 µg/mL Ca+2 and know or measure the density to be 1.033 g/mL then the ppm Ca+2 = (1000 µg/mL) / (1.033 g/mL) = 968 µg/g = 968 ppm.

When making dilutions the following equation is useful:

(mLA)(CA) = (mLB)(CB)

For example, to determine how much of a 1000 µg/mL solution of Ca+2 required to prepare 250 mL of a 0.3 µg/mL solution of Ca+2 we would use the above equation as follows:

(mLA)(1000 µg/mL) = (250 mL)(0.3 µg/mL)

(mLA) = [(250 mL)(0.3 µg/mL)]/ (1000 µg/mL)

(mLA) = 0.075 mL = 75 µL

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