Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: functions on February 17, 2012, 09:44:55 AM
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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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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+ + H2O :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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effervescence occur, a colourless gas evolved.
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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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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+ + H2O :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+