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I have read numerous online questions similar to the one I am about to ask. But for most of them the answer has been that the reason one can add water to the unknown concentration solution during a titration is because:

  1. Water has a pH of 7

  2. This is a volumetric analysis, so as long as one record the volume of the titrant and analyte used then there is no problem.

My questions are these:

Due to the effect of $\ce{CO2}$ in the atmosphere mixing with the distilled water, it is obvious that the pH of distilled water is on the acidic side (which we confirmed using a pH sensor). Therefore, why does the addition of distilled water not affect the pH? If it is acidic that should mean there are $\ce{H3O+}$ molecules in the distilled water when we add it into the analyte.

How do we take into account the imbalance of the moles of $\ce{H3O+}$ molecules versus $\ce{OH-}$ in each drop of the acidic distilled water (ex. pH of 5.6)

The only solution I have in mind right now is that all of these are acceptable uncertainties (such as the effect of the addition of an indicator), and that significant figures will wipe out these small errors.

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Unfortunately I could not find any sources that specifically address your question so I will be answering using logic. If you notice any fallacies or points you disagree with, please point them out.

  1. $\ce{CO2}$ dissolves in water at such low concentrations normally, the pH would not be greatly affected. Unless you are working in extremely small concentrations that require high accuracy, it would not be a very large problem. If you were, you could try to minimize the time the distilled water is opened, and perhaps work in a vacuum really quickly or find some way to remove that factor.

  2. If we were adding the distilled water to another solution, it is possible that the other solution was already in equilibrium with the partial pressure of $\ce{CO2}$ in the atmosphere and thus the concentration would remain constant ( a very small concentration ).

I do agree that protons may be added, but they would be so few, especially if you are reacting the reaction with acid or some strong base that it would be insignificant. Often times, conditions are not perfect, but good enough to be estimated and be close enough. However, measures can be taken to prevent your worries if the experiment demands it.

Any $\ce{CO2}$ probably already dissociated and is in equilibrium, or comes to it very closely. Opening up a coke and watching it enter equilibrium may serve as a nice visual. Don't worry about it in a titration unless you are trying to do something really really unusual. The $\ce{CO2}$ is insignificant in most titrations.

Note, the solubility is 1.45 g/L at 25 °C, 100 kPa max, so even a saturated solution would be not greatly affected. Furthermore, your solution is probably already in equilibrium so you have already accounted for the carbon dioxide. Think about how when you do a titration, it doesn't matter if you begin at 0 or whatever. As long as you account for it.

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  • $\begingroup$ I suppose my main question is that if I use an acidic distilled water in my lab, how could I account for it? Because I believe that in each drop of distilled water I use, I'm adding more H3O+ than OH- (which would not happen with a pH of 7 water). Therefore, it would take more base to neutralize my acidic solution of unknown concentration. $\endgroup$ – quidproquo Apr 16 '15 at 3:34

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