Dear Datta Kakade

realese of phosphorous is depends upon how much you taking in to the solution

for example:

1). KH2PO4; if you considering only about phosphorous for the compound. then.

if you want 1 gram of phosphate to be present in 1 liter of distilled water

* KH2PO4 molecular weight is 136.086 g/mol

PO4 atomic weight is 94.97 g/mol

then division of KH2PO4 Molecular weight/PO4 atomic weight

136.086/94.97 = 1.43 grams

so by dissolving 1.43 grams of KH2PO4 in 1 liter of distilled water gives 1 gram of concentration in 1 liter of distilled water.

2) in case of K2HPO4 molecular weight is 174.2 g/mol

PO4 atomic weight is 94.97 g/mol

then division of K2HPO4 Molecular weight/PO4 atomic weight

174.2/94.97 = 1.834 grams

so by dissolving 1.834 grams of K2HPO4 in 1 liter of distilled water gives 1 gram of concentration in 1 liter of distilled water.

Hope you get your answer.

all the best.

EXACTLY THE QUANTITY ADDED BY YOU WILL GOVERN THE RELEASE INTO WATER.OTHERWISE JAMEER HAS ALREADY EXPLAINED YOU BY MAKING CALCULATIONS PER GRAM OF PHOSPHORUS TAKEN.

Assuming that you are within the solubility limit, the whole amount added is in solution. Adjustments for ion activity can be made using speciation models. If there are other solutes, then you definitely need the models.

Dear JAMEER,

Actually I wanted to know there is no relationship of P with other compound with that one is bound like if we are adding urea as a source of N, then that will give the 46% N, like that I'm asking in case of KH2PO4 and K2HPO4.

I will provide the data about my experiment

we are adding 23.1 g of KH2PO4 and 4.3 g of K2HPO4 in 1 liter of distilled water respectively , so what was the exact amount of P in that 1 litre of distilled water?

Molecular weight of KH2PO4 is 136.08

It contains P equivalent to 22.78%

If you add 23.1 g of KH2PO4 in 1 L water, the exact amount of P in solution will be 5.26g

Molecular weight of K2HPO4 is 174.2

It contains P equivalent to 17.8%

If you add 4.3 g of K2HPO4 in 1 L water, the exact amount of P in solution will be 0.766g

This is in consideration if whole amount of P is soluble

thank you Sunita ma'am and JAMEER sir for your answers and suggestion on the to your information really help me lot.

Dear Jameer sir and Sunita Madam, I would like to preare phosphate (P) standard curve to estimate the orthophoshate/soluble P in water sample. But some research articles mentioned to take 4.394g of KH2PO4 in 1000ml of water which gives concentration of 1g/L. Can some body kindly explain about the calculation

Dear Mr. Datta

Molar mass of KH2PO4 is 136.085 g/mol

Molar mass of P is 30.974g/mol

Amount of KH2PO4 required to have 1g P/L is

136.085/30.974 =4.394

If you add 4.394g KH2PO4 in 1000 ml Volumetric flask and make the volume to 1L with water, you will get 1000 ppm P solution. Accordingly take 2 ml from the standard solution and make the volume to 100 ml to get 2 ppm solution. From this solution, prepare different concentration of P by taking 1, 2,4, 5,6 ml.

respected mam Sunita Gaind

i want to know about the calculation that among all these sources glucose, fructose maltose cellulose and sucrose

which one is the best carbon supplied source in culture media of 1 liter distilled water.

9

Dear Sunita Gaind

How can we prepare 50 ppm P stock solution from dipotassium hydrogen orthophosphate?

Molar mass of K2HPO4 is 174.2 g/mol

Molar mass of P is 30.974g/mol

Amount of K2HPO4 required to have 1g P/L is

174.2/30.974 =5.622

If you add 5.622g K2HPO4 in 1000 ml Volumetric flask and make the volume to 1L with water, you will get 1000 ppm P solution. For preparing 50 ppm P stock solution add 0.2811 g K2HPO4 in1000 ml volumetric flask and make the volume to 1 L

Datta Kakade , Sunita Gaind has clearly shown the calculations per g of KH2PO4 taken

Specifically, weight 0.2811 ±0.001 g KH2PO4 in small beaker, after 110℃ for drying 2 hours in oven and cooling down in a drying vessel, then add water to dissolve the chemical, and add 5 ml 1:1 H2SO4 before bring to the volume 1000 ml.

The 1:1 H2SO4 can help keep the stock solution for 6 months.

Dear Sunita Gaind ,

How can we prepare 100 ppm of phosphate stock solution from KH2PO4?

Dear Datta Kakade

Molar mass of KH2PO4 is 136 g/mol

Molar mass of P is 30.974 g/mol

Amount of KH2PO4 required to have 1g P/L is

136/30.974 =4.390

If you add 4.390 g KH2PO4 in 1000 ml volumetric flask and make the volume to 1L with water, you will get 1000 ppm P solution.

if you dissolve 0.439 g KH2PO4 (AR Grade) in de-ionized water and make the volume to 1 litre. you will get 100 ppm solution.

Dear Sunita Gaind :

How can I adjust my calculations considering how pure is the reagent that I am going to use?

Thank you very much.

Dear Gabriela,

Use reagent supplied by a standard company with mentioned purity. This is the standard calculation for release of P from KH2PO4/K2HPO4

Dear Mathieu Nsenga Kumwimba

Is there any specific reason why KH2PO4 must be drying 2 hours at 110℃? Thank You.

Dear Sunita:

How can I prepare 1000 ml of KH2PO4 solution with 5 mg P in each ml? Please!

Should we autoclave or filter-sterilize KH2PO4 for assays?