Question

The hydroxide ion concentration, [OH-], in each of your trials can be calculated from the concentration of the hydrochloric acid, 0.050 M HCl, and the volume of both the HCl and Ca(OH)2 used in the titration.

According to the net ionic equation (calcium hydroxide, Ca(OH)2 (s), in water) determine the ratio of moles of calcium ion, Ca2+, to moles of hydroxide ion, OH-, in the solution of calcium hydroxide, Ca(OH)2. Using this ratio, determine the molarity of calcium ion in the solution for each trial.

Calculate the average concentration for both the calcium ion, [Ca2+], and the hydroxide ion, [OH-]. Discuss the precision of your concentration values.

Data:

Trial 1 Trial 2 Trial 3

Ca(OH)2 0.87 mL 0.93 mL 0.91 mL

HCl 0.71 mL 0.73 mL 0.76 mL

Answer 1

Trial 1

moles of HCl = 0.05 M x 0.71 ml = 0.0355 mmoles

molarity of Ca(OH)2 = 0.0355 x 2/0.87 = 0.082 M

molarity of [Ca2+] = 0.082 M

moles of Ca2+/moles of OH- = 0.0355/0.071 = 0.5

Trial 2

moles of HCl = 0.05 M x 0.73 ml = 0.0365 mmoles

molarity of Ca(OH)2 = 0.0365 x 2/0.93 = 0.078 M

molarity of [Ca2+] = 0.078 M

moles of Ca2+/moles of OH- = 0.0365/0.073 = 0.5

Trial 3

moles of HCl = 0.05 M x 0.76 ml = 0.038 mmoles

molarity of Ca(OH)2 = 0.038 x 2/0.91 = 0.083 M

molarity of [Ca2+] = 0.083 M

moles of Ca2+/moles of OH- = 0.038/0.076 = 0.5

Average concentration of,

[Ca2+] = 0.081 M

[OH-] = 0.5 M

The results are within the experimental error and is good with precision for the given concentrations