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13 Dec 2019

The precipitation of aluminum hydroxide, Al(OH)3 (Ksp=1.3×10?33), is sometimes used to purify water.

At what pH will precipitation of Al(OH)3 begin if 6.80 lb of aluminum sulfate, Al2(SO4)3, is added to 1150 gallons of water (with a negligible change in volume)?

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Tod Thiel
Tod ThielLv2
17 Dec 2019

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Related questions

DATA FROM EXPERIMENT:

MASS OF WEIGHING PAPER = 0.3670 g
MASS OF WEIGHING PAPER + Al FOIL = 1.0000 g
MASS OF ALUM FOIL (NET WEIGHT OF ALUM) = 0.6330 g
MASS OF EMPTY 50 mL BEAKER = 29.3380 g
MASS OF BEAKER + ALUM CRYSTALS = 39.8840 g
MASS OF ALUM (NET WEIGHT OF ALUM) = 10.5460 g

we added about 35 mL of de-ionized water and 12mL of 6M KOHsolution to the beaker. I converted the molarity to moles and got6M KOH (.012 L) = .072 mol KOH

When metallic aluminum comes into contact with aqueous solutions ofstrong bases like potassium hydroxide, KOH, it reacts to formhydrogen gas and a salt containing both aluminum and potassiumions.

2 Al (s) + 2 KOH (aq) + 6 H2O (l) ? 2 KAl(OH)4 (aq) + 3 H2(g)

This salt, KAl(OH)4, is soluble in solutions of strong acids, suchas sulfuric acid, H2 SO4, and reacts to form aluminum hydroxide,Al(OH)3, and potassium sulfate, K2 SO4.

2 KAl(OH)4 (aq) + H2 SO4 (aq) ? 2 Al(OH)3 (s) + K2 SO4 (aq) + 2 H2O(l)

Excess sulfuric acid is required to dissolve the Al(OH)3, and toform aluminum sulfate, Al2 (SO4)3.

2 Al(OH)3 (s) + 3 H2 SO4 (aq) ? Al2 (SO4)3 (aq) + 6 H2O (l)

As this solution is cooled, the aluminum and potassium sulfatesalts crystallize out of the solution as the alum. The equationbelow shows only the ions and the water involved in thecrystallization process.

K+ (aq) + Al3+ (aq) + 2 SO4 2- (aq) + 12 H2O (l) ? KAl(SO4)2 * 12H2O (s)

MY QUESTION IS.. I DONT KNOW WHICH EQUATION TO USE TO CALCULATE THEPERCENT YIELD. SHOULD IT BE THE FIRST ONE SINCE IT IS BEFORE ANY OFTHE PROCEDURE WAS DONE? BUT FROM THE FIRST EQUATION, IT DID NOTPRODUCE THE ALUM CRYSTALS YET : KAl(SO4)2 * 12H20

EQUATION FOR PERCENT YIELD:
100 % x [mass of actual yield / mass of theoretical yield]

The Ksp of Al(OH)3 is 2 x 10-32. At what pH will a 0.8 M Al3+ solution begin to show precipitation of Al(OH)3?
A) 10.5
B) 6.0
C) 3.5
D) 3.0
E) 1.0

Aluminum metal will reduce water, but the reaction is extremely slow under neutral conditions at room temperature.

Al(s) + 3H2O(l) → Al(OH)3(s) + H2(g) very, very slow

In the presence of aqueous base, however, Al will readily react to form the very soluble Al(OH)4–1(aq) complex anion which can be neutralized with H2SO4(aq) in two steps. The first produces the very insoluble Al(OH)3(s). If the Al(OH)3(s) slurry is then immediately boiled with additional H2SO4(aq) the insoluble hydroxide will react completely to give a clear, colorless solution. After excess water is boiled off, cooling produces KAl(SO4)2.12H2O(s), Alum, as fine, white needles.

Al(s) + 3H2O + KOH ⟶ KAl(OH)4(aq) + H2(g) fast and exothermic at room temperature

KAl(OH)4(aq) + ½ H2SO4(aq) ⟶ Al(OH)3(s) + H2O(l) + ½ K2SO4(aq)

Al(OH)3(s) + H2SO4(aq) ⟶ ½ Al2(SO4)3(aq) + 3 H2O (requires heating)

½ Al2(SO4)3(aq) + ½ K2SO4(aq) ⟶ KAl(SO4)2(s)

Combining these four equations gives the overall reaction.

Al(s) + KOH(aq) + 2 H2SO4(aq) ⟶ [KAl(SO4)2(aq)] + H2O(l) + H2(g) and

[KAl(SO4)2(aq)] + 12 H2O(l) ⟶ KAl(SO4)2.12H2O(s) (Alum)

A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alum by following the experimental procedure given in this exercise. Assume that the Alum was 97.7% pure. Calculate the volume of 6.00 M H2SO4 (aq) needed just to precipitate Al(OH)3 (s).

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