How can you determine the molarity of a sodium hydroxide caustic?

Question

I have a 50% sodium hydroxide caustic. Is there a way to determine its molarity?

Answer 1

(Carefully!) weight out 0.80 g of this and dilute in water to 1 L volume. If the molarity is correct, the pH of the solution should be around 12.

Answer 2

Does 50% mean you have a 50% sodium hydroxide solution?
50% solution means there is 50g of sodium hydroxide solid resolved in 100g of water.

If the volume isn't changed during resolving, the molarity of sodium hydroxide is:
50g/(40g.mol-1x0.1L) =12.5 Mol/L

Actually, I guess the volume is changed during resolving. So, the precise method is:
1) weigh a cylindar
2) measure 10ml of the 50% solution by that cylindar
3) weigh the cylindar with 10ml solution
4) calculate the weight of 10ml 50% solution. Let's suppose it is W gram.
5) the molarity is:
W g/(3 x 40g.mol-1 x 0.01L) = (W /1.2) Mol/L

Answer 3

Titrate a sample against a standard acid (say 0.1 M) using phenolphthalien indicator to a colourless endpoint.

So
1. Weigh about 0.2 g (say w g)of your caustic in to conical flask.
2. Add a few drops of 1% phenolphthalien in 50% alcohol.
3. Titrate with 0.1 M hydrochloric acid to the colourless endpoint. Record mls used (say T ml). 0.1 M HCl should be a standard lab reagent.
4. Calculate NaOH %M/M as follows

NaOH %M/M = 4TM/w where

T Titre mls of HCl used
M Molarity of HCl ie 0.1M
w Weight g caustic taken g

EXTREME CAUTION

Samples of 50% caustic soda are particularly corrosive to skin. Take appropriate care and use gloves and eye protection when handling it.

Answer 4

The Balanced chemical equation is 2NaOH + H2SO4 = Na2SO4 + 2H2O enable first reactant = NaOH ( M1 , V1 , n1 are its respective molarity , volume and moles in contact ) 2d reactant = H2SO4 ( M2 , V2 , n2 are its respective molarity , volume , moles in contact ) M1 = 0.0253 M V1 = 25 mL = 0.0.5 L n1 = 2 ( The coefficient before NaOH ) V2 = 38.40 5 mL = 0.03845 enable molarity invovled of H2SO4 = M2 n2 = a million Neutralization - Molarity relation is M1V1 / n1 = M2V2 / n2 Filling respective values 0.0253 x 0.0.5 / 2 = M2 x 0.03845 / a million fixing for M2 M2 = 0.008224 M Molarity of Sulphuric acid answer (M2) = 0.008 M