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citric acid ionisation in water (1 Viewer)
- Thread starter shkspeare
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Xayma
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C<sub>6</sub>H<sub>8</sub>O<sub>7</sub>+3H<sub>2</sub>O < -------- > C<sub>6</sub>H<sub>5</sub>O<sub>7</sub><sup>3-</sup> + 3H<sub>3</sub>O<sup>+1</sup>
Thats for complete ionisation.
For the intermediates you can work out yourself.
Thats for complete ionisation.
For the intermediates you can work out yourself.
Last edited:
That should be 3H<sub>2</sub>O at the start, not 3H<sub>2</sub> .
Citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid), HOOC-CH<sub>2</sub>-C(OH)(COOH)-CH<sub>2</sub>-COOH is a triprotic acid, and like all triprotic acids, it ionises in a stepwise manner. ie:
C<sub>6</sub>H<sub>8</sub>O<sub>7 (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> C<sub>6</sub>H<sub>7</sub>O<sub>7</sub><sup>-</sup><sub> (aq)</sub> + H<sub>3</sub>O<sup>+</sup><sub> (aq)</sub>
C<sub>6</sub>H<sub>7</sub>O<sub>7</sub><sup>-</sup><sub> (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> C<sub>6</sub>H<sub>6</sub>O<sub>7</sub><sup>2-</sup><sub> (aq)</sub> + H<sub>3</sub>O<sup>+</sup><sub> (aq)</sub>
C<sub>6</sub>H<sub>6</sub>O<sub>7</sub><sup>2-</sup><sub> (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> C<sub>6</sub>H<sub>5</sub>O<sub>7</sub><sup>3-</sup><sub> (aq)</sub> + H<sub>3</sub>O<sup>+</sup><sub> (aq)</sub>
All three equilibria lie to the left, each further left than the one before - so, not many citric acidmolecules undergo the first ioinisation, and virtually none the third. In fact, many questions treat it as if it were a monoprotic acid - which is wrong, but you should recognise that it can happen, especially in assessments and trials, which may be written by people with not as much chemistry knowledge as would be desirable.
