Please fill out table... An experiment was performed to determine the K_sp of Ca

Question

Please fill out table...
An experiment was performed to determine the K_sp of Ca(OH)_2. A solution containing 2.00 x 10^-2M Ca(NO_3)_2 was saturated with Ca(OH)_2 and filtered. Two samples of the saturated solution were titrated with 5.021 x 10^-2M HCI, in order to determine the OH^-ion concentration in the saturated solution. In doing this problem, note that 1 mol of HCI reacts with 1 mol of OH^- ions. This relationship is the basis for the determination of the OH ion concentration: 1 mol Ca2^2+ ions are produced for every 2 mols of OH^- ions produced. The data resulting from this experiment are shown in Table 2. Table 2 Student data used to determine the K_sp of Ca(OH)_2. Calculate the following: number of moles of OH^- ions titrated, mol___[OH^-]_sat. sol'n___ {Ca^2+]_from dissolved Ca(OH)_2___[Ca^2+]_sat. sol'n___K_sp____average K_sp____

Explanation / Answer

0.033

Ksp = [Ca+2] [ OH-]^2 = 0.032 X (0.025)^2 = 0.00002

Trial 1 Trial 2 volume of saturated solution of Ca(OH)2 52.2 48.7 volume of 0.0521 M HCl 16.44 15.49 temperature 20.4 20.4 moles of OH- titrated = Moles of HCl millimoles of HCl = Molarity X volume 0.857 0.807 millimoles of OH- = 2*millimoles of HCl 1.713 1.614 [OH-] in saturated solution = moles / tot vol 0.025 0.025 [Ca+2] initially due to Ca(NO3)2 0.02 0.02 Ca+2 due to Ca(OH)2 = 1/2 [OH-] 0.012 0.013 Ca+ saturated solution = total concentration 0.032

0.033

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