Postlab8 - Moles of KOH Moles of Al(OH)3 = Moles of excess...

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1) Al(OH)3 is soluble in either acidic or basic solutions. Why is a basic, rather than an acidic, solution used dissolve the Al(OH)3 in this experiment? Dissolving Al(OH)3 in basic solution before adding an acid separates the two steps of dissolution and dissociation. If the Al(OH)3 was dissolved in an acidic solution, then at the same time as the dissolution the Al would dissociate from the OH, but the intermediate Al(OH)4- would not be formed when KOH was added and hydrolysis could take place on the Al3+. Dissolving in a basic solution makes sure all of the Al(OH)3 does not dissociate so that when one adds the KOH the Al(OH)3 becomes Al(OH)4- which is the needed intermediate for the alum reaction. Also, the precipitate that forms when the acid is added to the Al(OH)3 solution could interfere with the dissolution of the Al(OH)3. 2) Calculate the number of moles of excess hydroxide in the solution after all the Al(OH)3 has been converted to Al(OH)4-. 3.0g of Al(OH)3 (1 mol/78.00g) = .038 mol of Al(OH)3
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Unformatted text preview: Moles of KOH Moles of Al(OH)3 = Moles of excess hydroxide .05 mol of KOH .038 mol of Al(OH)3 = .012 mol of excess hydroxide 3) Describe the reaction between Al(OH)3 and OH- using Bronsted terminology. Bronstead Lowry acids are molecules that have an H+ molecule in them that dissociates when in solution. Bronstead Lowry bases are molecules that can accept OH- when in solution and have a lone pair of electrons. Al(OH)3 is BL acid and OH is BL base and the conjugate base of Al(OH)3. OH is added to the Al(OH)3 molecule making Al(OH)4-, this process decreases the amount of OH in solution which decreases the pH, which makes Al(OH)3 the acid and conjugate acid of OH. 5) What is the purpose of suspending the thread in the solution from which crystallization of alum is to take place? This is so that the crystals can have more surface area to form upon and it makes for easy removal of some of the crystals for weighing....
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