Question

Suppose a student adds NaOH too quickly, and surpasses the endpoint of the titration. a. Does this mean that too many moles or not enough moles of NaOH would be added to reach the endpoint? b.) How will this affect the perceived amount of acid? (i.e., will it be higher or lower than the actual number of moles of acid? c.) Will the calculated concentration of the acid be higher or lower than the actual value?

Answer 1

Ans. #1. If NaOH is added too quickly surpassing the endpoint, there has been excess (more that actually required) or too many NaOH has been added.

#2. Suppose there were X moles of a monoprotic acid (KHP), 25.0 mL in flask.

Case I: Correct or actual end point: The actual volumes of standardized NaOH solution (0.100 M) to be consumed to reach end point = 10.0 mL.

Moles of NaOH = Molarity x Vol. consumed in liters = 0.100 M x 0.010 L = 0.001 mol

Since 1 mol NaOH neutralizes 1 mol KHP, the number of moles of acid in flask is equal to the moles of NaOH.

So, you calculate that there are 0.001 moles of acid in the flask.

Now,

            Molarity of acid = moles / Volume in liters = 0.001 mol / 0.025 L = 0.04 M

So, actual [acid] = 0.04 M

Case II: Excess NaOH at end point: You dispensed 12.0 mL NaOH from burette instead of 10.0 mL.

Erroneous moles of NaOH consumed = 0.100 M x 0.012 L = 0.0012 mol

So, Erroneous moles of acid inn sample = 0.0012 mol

Now,

            [Acid] = 0.0012 mol / 0.025 L = 0.048 M

Note that erogenous [acid] is greater than actual value.

Conclusion: Therefore, if you dispensed extra (more than actual) volume of NaOH from the burette, the calculated [Acid] would be greater than the actual value because –

I. you added excess NaOH to neutralize the same number of moles of acid ; and

II. moles of NaOH consumed at endpoint is taken proportional to moles of acid in sample