__Altering the Concentration of Food Grade Hydrogen Peroxide____:__

__Altering the Concentration of Food Grade Hydrogen Peroxide__

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Firstly, let us try understand various ways in which concentration of substance/s are presented. The various ways to express concentrations are:

While preparing w/w solutions, weight of each constituent (Chemical) is used. Let us say a 100 Grams of Solution C is made by adding 10 Grams of Chemical A (Solute) to 90 Grams of Chemical B (Solvent). Now, this gives 10% w/w Solution of Chemical A.__% w/w (Percentage Weight per Weight):__Now, we use w/v calculations when a Solid Solute is dissolved into The Liquid Solvent. Now, assuming NaCl would completely dissolve into – within the molecules of Water, so, if I dissolve 10 Grams NaCl into 100 mL of water, I have 10% w/v solution of NaCl.__% w/v (Percentage Weight per Volume):____% v/v (Percentage Volume per Volume)__**:**Now, Mostly this measurement is used, when both ‘The Solute’ and ‘The Solvent’ are Liquids. Let us say,

- I add 30 mL of H2SO4 into 70 mL of Pure Distilled Water, I have 30 mL in a total volume of 100 mL; Thus I could call this solution at 30 % v/v.
- I add 750 mL Ethanol to 250 mL Pure Distilled Water (750 mL in total volume of 1000 mL), I have Ethanol Solution at 75 % v/v concentration.

__Few Miscellaneous Examples:__

__Volume per Volume To Weight per Volume:__

*I wish to convert “Ethanol Solution at 75 % v/v concentration” to “% w/v Concentration”:*

Density of Ethanol = 0.789 g/ml.

- Mass of 750 mL Ethanol = 750 * 0.789 = 591.75 grams
- 75 grams diluted in Pure Distilled Water to make 1000 mL Solution
- This Yields a Solution of Ethanol at 591.75/1000*100 (or 59.1 % w/v concentration).

- Now suppose I add 35 grams NaCl to Pure Water such as to make a total volume of 100 mL, the final NaCl solution would be 35 % w/v.

- If I dilute “100 mL Pure 35 % H2O2” by adding Pure Distilled Water to make a total volume of 1000 mL Solution, I then get H2O2 at a concentration of 10 % v/v.

__Now, let us go through two ways in which concentration of H2O2 is altered-diluted…__

__Now, let us go through two ways in which concentration of H2O2 is altered-diluted…____First Way:__

1 part 35% + 1 part water = 2 parts 17.5%

2 parts 17.5% + 2 parts water = 4 parts 8.75%

4 parts 8.75% + 4 parts water = 8 parts 4.375%

8 parts 4.375% + 8 parts water = 16 parts 2.1875%

*So we may say, to 100 mL 35% H2O2 we add 1500 mL to get 2.1875% H2O2*

__Second Way:__

Now, 50% Conc. H2O2 implies that in a 100 mL Solution, we have 50 mL H2O2, and 50 mL of Distilled Water.

Equation to use for concentration conversion is:

__[50 mL (Amount of H2O2 in 100 mL Solution–100 mL is amount taken at start)]__ / [(100 mL (Solution Volume at Start) + x (Water to Add to get Required Concentration)]** = Required Percentage of H2O2**

Let required amount of H2O2 to be added be x mL to get Required conc. of 3% Conc. of H2O2…

35/(100 + x) = 0.03 => x = 1066 mL which make a total solution of 1166 mL of 3% Concentration…

So, now let us consider an initial starting amount of 10 mL…and we have to convert 35% Conc. to 3% Conc. we go like this:

3.5/(10 + x) = 0.03 => x = 106.7 mL This is the amount of Pure Distilled water to be Added to 10 mL nitial Volume of 35% H2O2 Solution.