Answered

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Consider an acid-base titration in which the base is dispensed from a burette into a flask containing an acid. If any drops of the base adhere to the inner walls of the flask, but do not actually mix with the solution, the calculated acid concentration would be

Sagot :

Pstnonsonjokuidn

Answer:

Higher than the actual value

Explanation:

Titration is a volumetric process in which a known volume of solution is dispensed from a burette to react with a known volume of solution in a conical flask.

When acid-base titration is carried out in such a way that the base is in the burette and the acid is in the conical flask and drops of the base adhere to the inner walls of the flask, but do not actually mix with the solution, the calculated acid concentration would be higher than the actual value.

This is because;

From CA= CBVBnA/VAnB

When VB(volume of base) that reacted is lower than the actual volume recorded, then the calculated volume of CA(concentration of acid) is much higher than the actual value since drops of the base adhere to the inner walls of the flask.

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