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lorijessica16 · Apr 20, 2009, 05:58 AM

1) NaOH(s)--> Na+(aq) + OH- (aq)
2) Na+(aq) + OH(aq) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq)

Target
equation: NaOH(s) + H+(aq) + Cl(aq)--> H20 + Na+(aq) + Cl-(aq)

I need help solving this.. Thanks

Perito · Apr 20, 2009, 06:05 AM

Originally Posted by lorijessica16

1) NaOH(s)--> Na+(aq) + OH- (aq)
2) Na+(aq) + OH(aq) + H+(aq) + Cl-(aq) --> H2O + Na+(aq) + Cl-(aq)

Target
equation: NaOH(s) + H+(aq) + Cl-(aq)--> H20 + Na+(aq) + Cl-(aq)

Actually, it appears you have solved it.

$NaOH(s)\,-->\, Na^+(aq) + OH^- (aq)$

$HCl\, --> \,H^+\,+\,Cl^-$

$OH^-\, +\, H^+ \,=\, H_2O$

therefore, adding them together and eliminating redundancies on both sides of the "-->",

$NaOH\,+\,HCl\,-->\,H_2O\,+\,Na^+(aq)\,+\,Cl^-(aq)$

Depending on how you wish to express it, you can write it this way:

$NaOH\,+\,HCl\,-->\,H_2O\,+\,NaCl(aq)$

In this case, you don't emphasize that the NaCl is dissociated. If you wish to emphasize that NaOH and HCl are dissociated you can write this:

$Na^+\,+\,OH^-\,+\,H^+\,+\,Cl^-\,-->\,H_2O\,+\,Na^+(aq)\,+\,Cl^-(aq)$

But the Na+ and Cl- on both sides of the equation is sort of silly. They're simply "spectator ions".

$NaOH\, +\, HCl\, =\, Na^+\,+\,OH^-\,+\,H^+\,+\,Cl^-\,-->\,H_2O\,+\,Na^+(aq)\,+\,Cl^-(aq)$

I'm not sure how this relates to Hess's law which is about energy changes.