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?
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.
To determine the molarity of Ba(OH)2 x 8H2O in the saturated solution, we can use the neutralization reaction between Ba(OH)2 and HCl.
The balanced neutralization reaction between Ba(OH)2 and HCl is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
We know that 31.24 mL of 0.1250 M HCl was required to neutralize the hydroxide ion in the saturated solution.
Step 1: Determine the number of moles of HCl used.
To do this, we use the formula: moles = volume (in L) x concentration (in mol/L).
moles HCl = 31.24 mL × (1 L / 1000 mL) × (0.1250 mol/L) = 0.003905 mol HCl
Step 2: Use stoichiometry to find the moles of Ba(OH)2 reacted with HCl.
From the balanced equation, we see that 1 mol of Ba(OH)2 reacts with 2 mol of HCl.
moles Ba(OH)2 = 0.003905 mol HCl × (1 mol Ba(OH)2 / 2 mol HCl) = 0.001953 mol Ba(OH)2
Step 3: Calculate the volume of the sample used.
Given that the sample used is 10.00 mL, we have:
volume (in L) = 10.00 mL × (1 L / 1000 mL) = 0.0100 L
Step 4: Calculate the molarity of Ba(OH)2 x 8H2O.
Using the formula: Molarity = moles / volume (in L), we have:
Molarity Ba(OH)2 = 0.001953 mol Ba(OH)2 / 0.0100 L = 0.1953 M
Therefore, the molarity of Ba(OH)2 x 8H2O in the saturated solution is 0.1953 M.