Caustic soda 100% to Baume conversion?

Given say needing to input 0,5 gram per litre of Caustic Soda 100% strength how would I convert this if I know our Caustic soda is 46°Be (Baume)?

i.e. I want to convert from a percentage value to Baume.

Given this is Caustic soda I would guess that 100% means solid NaOH and I know 46°Be is a Lye.

I've gone in circles with SG tables and am finding nothing in this forum or elsewhere online.

TIA

[Mrs X]

According to www.chembuddy.com 5%w/w NaOH has a density of 1.0554g/mL.

If you go by the Baume calculation on wikipedia,

For liquids heavier than water: s.g. = 145 ÷ (145 - degrees Baumé)

Thus: 1.0554=145 ÷ (145 - °Be)

Rearranged: °Be = -((145/1.0554) -145) = 7.6

For 46°Be, you would need a solution density of 145 ÷ (145 - 46) = 1.46g/mL. According to the chembuddy table, your solution would be about 43%w/w NaOH. This is 43g NaOH in 57g of water. The solution needs to be kept at room temp. You could weigh the clean dry flask beforehand, and weigh it again after you have finished to check that the density is correct.

Might pay to find a reference from another source as well, as some things on Wikipedia are opinion only.

I wish to reduce 20 deg Baume Muriatic Acid(31.45%) to 10%. I need 16oz of the 10% solution.

[JohnS]

Originally Posted by matawa28@att.net

I wish to reduce 20 deg Baume Muriatic Acid(31.45%) to 10%. I need 16oz of the 10% solution.

Muriatic acid is hydrochloric acid. Using table 24 of "Concentrative Properties of Aqueous Solutions" from the Handbook of Chemistry and Physics, and
20 °Baume = SG of 1.16, This interpolates to 31.72% w/w (slightly different from your number) and solute of 367.37 g/L

You desire a 10% solution which is a table entry and solute is 104.8 g/L. 0.5 L requires 52.4 g
Use 52.4 g x 1000 mL/367.37 g = 142.6 mL and dilute to final volume of 500 mL.

If you have to do many of these calculations, I suggest acquiring HC&P. The new editions are rather expensive, recent prior editions are available on eBay at a good discount (that's how I got mine).

Please note you are dealing with VERY strong initial solutions and pretty strong diluted solutions. Please observe proper precautions.

How many liters of a 25% caustic solution would be needed to equal 8# of dry caustic?

[JohnS]

Originally Posted by Unregistered

How many liters of a 25% caustic solution would be needed to equal 8# of dry caustic?

8 lb = 3.6288 kg
if solution is 25% w/w, you need 3.6288 kg/0.25 = 14.515 kg

My HC&P says the density of 25% NaOH is 1.274 kg/L, so
14.515 kg x 1 L/1.274 kg = 11.39 L

what is the amount of caustic soda 99% required to put into 1 ltr of water to make 50deg Be? help me pls , thank you

[JohnS]

Originally Posted by Unregistered

what is the amount of caustic soda 99% required to put into 1 ltr of water to make 50deg Be? help me pls , thank you

See coversion to SG in post 2 50° Baume is SG = 1.52632
Density of water at 20 °C is 0.998 kg/L so density is 1.52327 kg/L.

Using that density and this calculator http://www.handymath.com/cgi-bin/spc...i?submit=Entry

solution is 49.784% NaOH ( and 50.216% water)

You want an unknown weight of solution of which 50.216% is 0.998 kg (1 L) of water or 0.998/0.50216 = 1.98741 kg of solution.
The solution is 49.784% NaOH or 0.98941 kg, but is only 99% pure
0.98941 kg/0.99 = 0.99941 kg
Unless you can measure this precisely, 1 kg of 99% caustic soda and 1 L water is pretty close.