How to prepare ppm solutions.?

How to prepare ppm solutions.?
Give me a example.

ppm means parts per million.
When it prepare for given method.

1mg in 1000ml =1ppm solution

OK, but weight is varying based on potency of the material. eg. potency is 98% weight is 1*100/98 = 1.02 mg/ltr. is equal to 1 ppm solution.

Thanks

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PPM = PARTS PER MILLION

PPM is a term used in chemistry to denote a very, very low concentration of a solution. One gram in 1000 ml is 1000 ppm and one thousandth of a gram (0.001g) in 1000 ml is one ppm.

One thousanth of a gram is one milligram and 1000 ml is one liter, so that 1 ppm = 1 mg per liter = mg/Liter.
PPM is derived from the fact that the density of water is taken as 1kg/L = 1,000,000 mg/L, and 1mg/L is 1mg/1,000,000mg or one part in one million.

OBSERVE THE FOLLOWING UNITS

1 ppm = 1mg/l = 1ug /ml = 1000ug/L
ppm = ug/g =ug/ml = ng/mg = pg/ug = 10 -6
ppm = mg/litres of water

1 gram pure element disolved in 1000ml = 1000 ppm

PPB = Parts per billion = ug/L = ng/g = ng/ml = pg/mg = 10 -9

MAKING UP 1000 PPM SOLUTIONS

1. From the pure metal : weigh out accurately 1.000g of metal, dissolve in 1 : 1 conc. nitric or hydrochloric acid, and make up to the mark in 1 liter volume deionised water.

2. From a salt of the metal :
e.g. Make a 1000 ppm standard of Na using the salt NaCl.

FW of salt = 58.44g.
At. wt. of Na = 23
1g Na in relation to FW of salt = 58.44 / 23 = 2.542g.
Hence, weigh out 2.542g NaCl and dissolve in 1 liter volume to make a 1000 ppm Na standard.

1. From an acidic radical of the salt :
e.g. Make a 1000 ppm phosphate standard using the salt KH2PO4

FW of salt = 136.09
FW of radical PO4 = 95
1g PO4 in relation to FW of salt = 136.09 / 95 = 1.432g.
Hence, weigh out 1.432g KH2PO4 and dissolve in 1 liter volume to make a 1000 ppm PO4 standard.
You can consider potency or Assay content for above calculations by applying correction factor as follows,

Actaul Qty. of solid to be taken to prepare the solution of desired strength =
Qt. obtained therotically X 100 / (Potency or Assay in % )

Hope I have explained you to understand the concept and calculations.

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