In an experiment to measure the solubility of Ba(OH)2 x 8H2O in water, an excess of solid Ba(OH)2 x 8H2O was stirred with water overnight. the saturated solution was allowed to stand until all the undissolved solid had settled. a 10.00mL sample of the clear supernatant liquid was titrated with 0.1250 M HCl. If 31.24 mL of 0.1250 M HCl was required to neutralize the hydroxide ion in the saturated solutions, what was the molarity of Ba(OH)2 x 8H2O in the saturated solution?

Question

In an experiment to measure the solubility of Ba(OH)2 x 8H2O in water, an excess of solid Ba(OH)2 x 8H2O was stirred with water overnight. the saturated solution was allowed to stand until all the undissolved solid had settled. a 10.00mL sample of the clear supernatant liquid was titrated with 0.1250 M HCl. If 31.24 mL of 0.1250 M HCl was required to neutralize the hydroxide ion in the saturated solutions, what was the molarity of Ba(OH)2 x 8H2O  in the saturated solution?

DrBob222 · Answer

I will omit the H2O until the final step.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O

mols HCl = M x L = ? in the 10 mL sample.
mols Ba(OH)2 = 1/2 that in the 10 mL sample.
mols in a L sample = ? mols in the 10 mL sample x (1000/10) = x mol.
M = mols/L.