For a.
eqn 1. mols NaOH = M NaOH x L NaOH
eqn 2. mols acid = mols NaOH
eqn 3. M acid = mols acid/L acid
You're right, the NaOH is diluted which means you will add MORE NaOH when you titrate the acid. Look at eqn 1, If you add more volume you THINK you have more mols NaOH, that leads to more mols acid from eqn 2 and that increases the M of the acid from eqn 3.
Didn't Devron answer this for you a day or so ago?
For b. If a wet flask is used for the titration the M of the acid is unchanged. It's true you dilute the acid when you add it but it's also true yu dilute the base when you titrate. The water doesn't affect mols acid used or mols NaOH used.
c. If an air bubble in the buret is released during the titration it means the buret reading is too high because you've measured the volume of the base used as well as the volume of the air bubble. Now follow that from eqn 1, 2, 3.
- What effect would be on the normality of unknown acid? Explain why. (standarized
NaOH in the burette)
a) if a wet burette is used for NaOH (that has a known normality)
b) if a wet conical flask is used for the titration against the acid
c) if an air bubble is out from the burette
i know for A that water will dilute the base causing an error in calculations of the unknown acid concentration .. i cant figure the effect on normality though ?
1 answer