You can find other answers to the CO2 question and some analytical chemists don't agree with this but I think it may or may not make a difference. When CO2 reacts with NaOH it forms Na2CO3.
CO2 + 2NaOH ==> Na2CO3 + H2O. For every mole CO2 added we form 1 mol Na2CO3 and use up 2 mols NaOH. No one disagrees on that step. I think that when titrating the NaOH, IF we use methyl red as an indicator we titrate CO3^2- all the way (meaning CO3^2- + 2H^+ ==> H2CO3 ==> CO2 + H2O). In this way all of the CO3^2- formed by the reaction with 2NaOH is titrated so there is no difference in the M NaOH this way vs no dissolved CO2. If, however, we use phenolphthalein as an indicator, only one of H is added (H^+ + CO3^2- ==> HCO3^-) and there IS a difference. Less HCl is used in the titration step which makes mols NaOH present less than it should be which makes M NaOH too low. When standardizing NaOH I ALWAYS use methyl red or another appropriate indicator to avoid this problem.
In this experiment NaOH is standardized to titrate it with vinegar so thtat the percent by mass of the acetic acid can be determined.
How does dissolved CO2 in distilled water affect the accuracy of the determination of a NaOH solution's concentration? How is the term acid anhydride applied here?
Is the answer that the added CO2 reacts with NaOH to form Na2CO3 when titrated and the NaOH and the Na2CO3 is also titrated, so there is no difference?
Or is there a difference? high determination of a NaOH solution's concentration? low? How does the CO2 affect the determination of a NaOH solution's concentration?
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