The reaction is the following:
Mg(OH)2 + 2 C6H5COOH = Mg(C6H5COO)2 + 2 H2O
benzoic acid + Sodium Hydroxide ==> water + sodium benzoate
The reaction between hydrofluoric acid (HF) and barium hydroxide (Ba(OH)2) would produce barium fluoride (BaF2) and water (H2O).
Barium hydroxide, Ba(OH)2, is a base.
White precipitate will formed which is barium sulfate.
yes.
benzoic acid + Sodium Hydroxide ==> water + sodium benzoate
The reaction between hydrofluoric acid (HF) and barium hydroxide (Ba(OH)2) would produce barium fluoride (BaF2) and water (H2O).
Barium hydroxide, Ba(OH)2, is a base.
Ba(OH)2 + H2SO4 ---> BaSO4 + 2H2O So one molecule of sulphuric acid will be neutralized by one molecule of barium hydroxide. Or two molecules of sulphuric acid will be neutralized by two molecules of barium hydroxide.
White precipitate will formed which is barium sulfate.
yes.
Barium sulfate is the precipitate
When barium hydroxide solution is mixed with sulfuric acid, a white precipitate of barium sulfate is formed due to a double displacement reaction. Barium sulfate is insoluble in water and thus forms a solid precipitate. This reaction is used in chemistry labs to test for the presence of sulfate ions.
To find the molarity of the barium hydroxide solution, first calculate the number of moles of hydrochloric acid used in the titration. Then use the stoichiometry of the reaction to determine the number of moles of barium hydroxide present. Finally, divide the moles of barium hydroxide by the volume of the solution in liters to get the molarity.
Barium hydroxide, Ba(OH)2 is a strong base.
Barium fluoride can be considered to be the salt of barium hydroxide (a weak base) and HF (a strong acid). And a solution of BaF2 will be weakly acidic.
Ba is Barium, OH is Hydroxide and the 8 H2O is Octahydrate. This results in Barium Hydroxide Octahydrate.