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Author Topic: sodium alum in baking powder  (Read 965 times)

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functions

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sodium alum in baking powder
« on: February 17, 2012, 03:44:55 AM »

1. I dissolved about 0.5g of sodium alum,NaAl(SO4)2.12H2O in water. I then measured the pH of the solution I get pH 3. My lab experiment ask 'Write an equation that accounts for the formation of the hydrogen ions from this substance'.

In this case can I write

Al(H2O)6+++ + H2O ::equil::H3O+ + Al(H2O)5OH++
Al(H2O)5OH++ +H2O ::equil::H3O+ + Al(H2O)4(OH)2+
Al(H2O)4(OH)2+ +H2O ::equil::H3O+ +Al(H2O)3(OH)3
                                ::equil::Al(OH)3 + 3H2O + H3O+
All of these equation involve only H3O+ ion not H+ ion, but the question ask for H+ ion. Does anybody know how to express them in H+ ion?

2.Mix about 1.5g of finely powdered NaAl(SO4)2.12H2O with about 0.5g of dry sodium bicarbonate in a 100-mL beaker. Add 5mL of water to the mixture.My lab experiment ask 'Is there any evidence of a reaction?', 'Write an equation that account for the reaction that occured'.     

I have no idea with this question.                               
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fledarmus

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Re: sodium alum in baking powder
« Reply #1 on: February 17, 2012, 04:07:53 AM »

For the first question, you're dealing with two different pedagogical generations of understanding acid solutions in water. A long time ago, when I was taking general chemistry, dirt was old, and dinosaurs were just being invented, acidity was expressed in terms of [H+] in water. Shortly thereafter, with the argument that free protons couldn't possibly exist in water, acidity began to be expressed as [H3O+], drived from the formalistic reaction

H+ + H2:rarrow: H3O+

In simplistic terms, "hydrogen ions" and "hydronium ions" in a water solution are the same thing; your reactions are probably sufficient.

For the second question, start with the first part - what did you observe when you added the two powders together?
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functions

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Re: sodium alum in baking powder
« Reply #2 on: February 17, 2012, 05:34:37 AM »

effervescence occur, a colourless gas evolved.
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Arkcon

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Re: sodium alum in baking powder
« Reply #3 on: February 17, 2012, 08:42:05 AM »

effervescence occur, a colourless gas evolved.

Good.  Now, would you say that was a reaction?  If you want to predict a chemical equation, you might want to know some other properties of the gas.
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functions

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Re: sodium alum in baking powder
« Reply #4 on: February 17, 2012, 04:08:34 PM »

test the gas with a burning splint. How could we know there is a reaction instead of change of physical state of the reactants? Also, can you explain why the last equation of the equations

Al(H2O)6+++ + H2O ::equil::H3O++ Al(H2O)5OH++
Al(H2O)5OH++ +H2O ::equil::H3O+ + Al(H2O)4(OH)2+
Al(H2O)4(OH)2+ +H2O ::equil::H3O+ +Al(H2O)3(OH)3
                                ::equil::Al(OH)3 + 3H2O + H3O+

is written halfway only? Actually I copy these equation from a book. Does it mean
Al(H2O)4(OH)2+ +H2O ::equil::H3O+ +Al(H2O)3(OH)3 and Al(H2O)4(OH)2+ +H2O ::equil::Al(OH)3 + 3H2O + H3O+ are the same equation?

since you said

For the first question, you're dealing with two different pedagogical generations of understanding acid solutions in water. A long time ago, when I was taking general chemistry, dirt was old, and dinosaurs were just being invented, acidity was expressed in terms of [H+] in water. Shortly thereafter, with the argument that free protons couldn't possibly exist in water, acidity began to be expressed as [H3O+], drived from the formalistic reaction

H+ + H2:rarrow: H3O+

In simplistic terms, "hydrogen ions" and "hydronium ions" in a water solution are the same thing; your reactions are probably sufficient.

For the second question, start with the first part - what did you observe when you added the two powders together?

Is the three equations is eqivalent to the following equations

Al(H2O)6+++::equil::H++ Al(H2O)5OH++
Al(H2O)5OH++::equil::H+ + Al(H2O)4(OH)2+
Al(H2O)4(OH)2+::equil::H+ +Al(H2O)3(OH)3
                      ::equil::Al(OH)3 + 3H2O + H+

Can I combine these equations by writing

Al(H2O)63+::equil::Al(OH)3 + 3H2O + 3H+
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