Question

"In the weak acid titration, if you had not prepared your buret properly and a considerable amount of water was left in it, what would have been the effect on the calculated value of Ka?"

I generally know that the Ka of an acid remains constant no matter what, but in this case, the Ka of the weak acid being titrated can be determined and is equal to the pH at half equivalence. How would a more dilute strong base have an effect on the pH at half equivalence of the weakly acidic solution? Knowing the effect on ph, I assume we can deduce the effect on pKa and thus on the Ka of the acid.

Answer 1

Ans. For simplicity of expression, we proceed with a monoprotic, weak acid taken in the conical flask is being titrated with strong base in burette.

# Let’s say there are 0.05 moles of the acid in flask. The volume of flask does not matter because adding water to the flask does not alter the number moles of acid (though concertation, i.e. number of moles per unit volume is decreased).

So, we are going to titrate 0.05 moles of a weak monoprotic acid.

# Let’s say, the molarity of NaOH in the burette is 2.0 M.

1 mol NaOH neutralizes 1 mol weak monoprotic acid.

So, required moles of NaOH to reach endpoint = 0.05 mol

Required volume of NaOH in burette = Required moles / Molarity of NaOH

                                                = 0.05 mol / (2.0 M)                         ; [1M = 1 mol/ L]

                                                = 0.05 mol / (2.0 mol/ L)

                                                = 0.025 L

                                                = 25.0 mL

So, if you have conditioned your burette correctly, 25.0 mL of NaOH shall be consumed to reach the end point.

# Required volume of NaOH when water is present as contaminant in burette.

When water is present, the NaOH solution will be diluted.

So, you required more volume of NaOH to neutralize the same acid.

Say, due to dilution, 30.0 mL of NaOH is consumed to reach the endpoint.

# Affect on Strength of Acid: Unaware of dilution, you noted that 30.0 mL of NaOH is consumed.

Now, if you calculated the moles of NaOH consumed you get 0.6 moles. So, you assume there are 6.0 moles of acid in the flask. Or, the calculated concertation of acid is greater than the actual one.

# Effect on pKa: How to calculate the pKa value? Do you really account the moles of base consumed?

The pKa of weak acid is approximated using the titration curve, but using the moles of base consumed.

A graph of pH vs volume of base consumed is plotted.

The midpoint of horizontal line when pH change in very low compared to volume of base added is take pKa value. We don’t even need the actual equivalence point if we know the acid to be monoprotic- calculating the equivalence point is of no use for approximating the pKa value from graph.

Since the base is diluted, the horizontal region depicting the pKa value shifts to the right on X-axis (as more NaOH is consumed) but the pH on Y-axis remains the same.

So, there would be no effect on the pKa value irrespective of the strength and volume of base consumed to reach the endpoint.