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If I do the following reaction: $$\ce{Na3PO4 + 3HNO3 -> 3NaNO3 + H3PO4}$$ Does nitric acid completely react with the Trisodium phosphate to form a weak acid and a salt, or does it form a small amount of the weak acid or nothing? If it does form the weak acid, will this reaction create heat? I have only really dealt with neutralization reaction so I don't really understand what will happen, any help would be greatly appreciated.

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This would be a simpler question to answer for a salt of a monobasic acid, like sodium acetate. But, for the specific reaction you ask about:

First, we must consider the conjugate acid of trisodium phosphate: phosphoric acid, which will be formed when nitric acid is added. It has three $\mathrm{p}K_\mathrm{a}$ values for the three decreasingly acidic hydrogens: $2.15,$ $7.20,$ and $12.32.$ These values are measured in dilute water solutions, so they do not apply exactly to a reaction between pure nitric acid and pure trisodium phosphate. But these values do indicate that, by combining pure trisodium phosphate with three equivalents of a strong acid like nitric acid $(\mathrm{p}K_\mathrm{a} < 0)$ you would effectively have a complete conversion to phosphoric acid and sodium nitrate. And, yes, the reaction will be exothermic.

That was the simple answer. But if the reaction is carried out in water solution, which would help to keep the reaction from overheating and allow control of the rate of combining the reactants, we note that the first $\mathrm{p}K_\mathrm{a}$ of phosphoric acid is only $2.15$ units higher than that of nitric acid in water (effectively, hydronium ion with a $\mathrm{p}K_\mathrm{a}$ of $0)$ — in other words, phosphoric acid is $10^{-2.15}$ or about $1\%$ as strong an acid as hydronium ion.

So, after the reaction is complete, we would have about $99\%$ of the phosphate in the form of phosphoric acid and about $1\%$ in the form of the dihydrogen phosphate anion. There would also be small amounts present of monohydrogen phosphate dianion and phosphate trianion, whose concentrations can calculated as $10^{-\mathrm{p}K_\mathrm{a}}$ for the other two $\mathrm{p}K_\mathrm{a}$ values. So, technically the reaction will be about $99\%$ "complete".

Meanwhile, keep in mind that the sodium ions are "observer" species that do not participate in the acid-base reaction.

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