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Title: What does 2H2O part of H2C2O4*2H2O mean and how do I write a balanced equation using oxalic acid dihydrate? Post by: iliketorifk on Feb 15, 2013 I don't understand how the 2H2O behaves when writing a chemical equation of oxalic acid dihydate dissolved in water and then titrated with NaOH to the end point. I'm having difficulty writing a balanced equation to eventually solve for molarity.
Title: Re: What does 2H2O part of H2C2O4*2H2O mean and how do I write a balanced equation using oxalic acid Post by: acasto0871 on Feb 18, 2013 The dihydrate part of the reactant does not take part in the reaction. Throughout my training I was never taught to include it in any balanced equation. I do not think that students would be taught to do so today:
the balanced equation is: (COOH)2 + 2NaOH → (COONa)2 + 2H2O If you insist on writing oxalic acid as H2C2O4: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O And if you insist on including the dihydrate: H2C2O4.2H2O + 2NaOH → Na2C2O4 + 4H2O The problem with this last equation is that the reaction is carried out in aqueous medium, and the water of crystalisation or hydration of the oxalic acid is no longer associated with the oxalic acid when it is in solution. secondly: The product sodium oxalate occurs as a trihydrate, so if you have to consider the oxalic acid as a dihydrate, then in the equation you should consider that the sodium oxalate is a trihydrate, and you would get: H2C2O4.2H2O + 2NaOH → Na2C2O4.3H2O + H2O - which I think is actually quite ridiculous. Title: Re: What does 2H2O part of H2C2O4*2H2O mean and how do I write a balanced equation using oxalic acid Post by: sarah.brush on Feb 18, 2013 H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
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