..what effect on the molarity of NaOH and reported value of unknown strong acid if a drop of NaOH add here to the side of Erlenmeyer flask during the standardization of NaOH solution..
The part I have highlighted is not clear. However assuming it means that some NaOH spills out of the flask (and assuming there were 25cm3 of NaOH in the flask); then the volume of NaOH would be less than 25cm3, and therefore less acid will be required to react with it. The calculated molarity of the acid would be more than the actual, and consequently the calculated molarity of NaOH from the stoichiometric equation would be less than the actual molarity.
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