Chem 1212 Lab Manual-Revised 11206 Molarity of HCI-05\'s Y4M *Trial #1 Trial #2

Question

Chem 1212 Lab Manual-Revised 11206 Molarity of HCI-05's Y4M *Trial #1 Trial #2 Volume ofCa(OH)2- 10-Bet/- Initial Buret Reading-seeeme- Final Buret Reading -42SeeL Volume HCI Delivered _ 504L Solubility (M) Kae of Ca(OH) io.com 0p) 14 pts) 4gts 14 pts) Sample calculation of solubility (M) of Ca(OH) in deionized H:O, Trial 1.p 2. * Sample Calculation of K-p. Trial 1(3 Average Solubility (M) Si Average Kap Average Solubility (g100ml.) (3m) Average pKap Calculation of average solubility of Ca(OH): in g/ 100 mL

Explanation / Answer

First we have to write the chemical equation for the titration of Ca(OH)2 with HCl

Ca(OH)2 + 2HCl ==== CaCl2 + 2 H2O

moles of HCl required = Molarity * Volume

moles of HCl = 0.05344 * 0.0075 = 0.0004 required, the chemical equation shows that we need 1 mole of Ca(OH)2 to neutralize 2 moles of HCl

moles of Ca(OH)2 = 0.0004 / 2 = 0.0002 moles

Volume of Ca(OH)2 = 0.01 Liters = 10 ml

Molarity = 0.0002 / 0.01 = 0.02 = s = molar solubility

Ca(OH)2 === Ca+2 + 2OH-

........................s...........2s

Kps = s * (2s)2

Kps = 4s3 = 4 * 0.023 = 0.000032 = 3.2 x10-5

For trial 2

moles of HCl = 0.05344 * 0.007 = 0.000374 moles of HCl

moles of Ca(OH)2 =  0.000374 / 2 = 0.000187

volume Ca(OH)2 = 10 ml = 0.01 L

Molarity of Ca(OH)2 = 0.000187 / 0.01 = 0.018704 M

Kps = 2.617 x 10-5

Average Kps = 2.9183 x 10-5

Average pKps = -log ( 2.9183 x 10-5 ) = 4.53

Average solubility = 0.01937 M

0.01937 M = 0.01937 mol / L

if we multiply this by the molar mass of Ca(OH)2 = 74 g/gmol

0.01937 mol / L * 74 g/gmol = 1.433 g/L

This is the ammount of Ca(OH)2 in one liter of solution, 1 Liter is 1000 ml, we only want to know about 100 ml, so we just need to divide by 10

1.433 / 10 = 0.1433 g / 100 ml

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