Titration - Neutralization Of Hydrobromic Acid By Calcium Hydroxide
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Titration - Neutralization Of Hydrobromic Acid By Calcium Hydroxide Titration is the process whereby a solution of known concentration called the standard solution is combined with a solution of unknown concentration in order to determine the unknown concentration or the quantity of solute in the unknown. An indicator is used to show the end point of the titration. This end point coincides with the equivalent point, the point at which equivalent quantities of reactants are brought together stoichiometrically.

A sample of solid Ca(OH)2 is stirred in water at 300C until the solution contains as much dissolved Ca(OH)2 as it can hold. A 100-mL sample of this solution is withdrawn and titrated with 5 x 10-2 M HBr. It requires 48.8 mL of the acid solution for neutralization.
What is the molarity of the Ca(OH)2 solution?
What is the solubility of Ca(OH)2 in water, at 300C in grams of Ca(OH)2 per 100 mL of solution?

The strings: S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).

The math:
Although we are calculating, mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type interaction (chemical). (a) Chemical equation:
2HBr + Ca(OH)2 -------> CaBr2 + 2H2O
0.05 mol in 1 Liter of HBr (concentration of HBr)
So, (0.05)(0.0488) = 0.00244 mol in 48.8 mL
Mole ratio HBr:Ca(OH)2 = 2:1 (from balanced equation)
So, mol of Ca(OH)2 = 0.00244/2 = 0.00122 mol
So, 0.00122 mol of Ca(OH)2 in 0.100 L
So, 0.00122/0.100 = 0.0122 in 1 L = 0.0122 M = 1.22 x 10-2 M.

(b)
Molar mass of Ca(OH)2 = 40 + 32 + 2 = 74 g
So, mass of Ca(OH)2 in 100 mL = 0.00122 x 40 = 0.0903 g
So, solubility of Ca(OH)2 in 100 mL = 0.0903 g.

Math

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