Question

Comparing the two titration curves (strong-strong titration vs. strong-weak titration), what is the difference regarding the pH change in the different regions? Explain why there is this kind of difference in terms of existing ions in the system and possible (equilibrium) reactions Saved Normal . BIO X1 X1 EEE fxl le BELI 9 illi

Answer 1

A titration curve is a sigmoidal curve (i.e., looks like an S). A generic titration curve is shown below.

After Equivalence Point Equivalence Point PH Before Equivalence Point 5 10 15 20 25 30 35 40 Volume of NaOH added

As shown in the picture, the steep straight line in the midde of the curve indicates the pH change at the equivalence point and the tail of the S-curve is before the equivalence has reached and the head (on far right) is after the equivalence has reached.

For strong-strong titration: Let us take an example of HCl vs NaOH titration.

1. Before equivalence point:

i. Before adding any NaOH:The sample solution contains only HCl, a strong acid. It will produce H3O+ ions in the medium and therefore, the pH of the medium will be very low, pH $\sim$ 1 - 3 (highly acidic medium).

Reaction: $HCl + H_2O \rightarrow Cl^- + H_3O^+$

Therefore, the tail of the curve (almost flat portion to the left) will be before the pH = 7, i.e. $\sim$ 1 - 3.

ii. On adding NaOH: The NaOH will start to react with HCl resulting in decrease of H3O+ ion concentration, increasing the pH of the medium.

Reaction occurring during titration: $HCl + NaOH \rightarrow NaCl + H_2O$

2. At the equivalence point: All of the HCl has been neutralized and hence only NaCl salt and water will be present in the medium. Since NaCl is a salt made from a strong acid and a strong base, it will not undergo hydrolysis and hence the pH of the medium will be that of water (neutral), i.e. pH = 7. Therefore, the steep rise in the pH vs volume of NaOH added curve when equivalence has reached will be around the pH = 7.

3. After the equivalence point: After all of the HCl has been consumed by NaOH, addition of excess NaOH would result in excess of OH- ions and hence a basic medium, pH $\sim$ 12-13 (strongly basic medium). Therefore, the head of the curve (almost flat portion to the right) will be after the pH = 7, i.e. $\sim$ 12-13 .

For strong-weak titration: Let us take an example of Acetic acid (CH3COOH) vs NaOH titration.

1.Before equivalence point:

i. Before adding any NaOH: The sample solution contains only CH3COOH, a weak acid. It will produce H3O+ ions in the medium and therefore, the pH of the medium will be acidic,i.e. pH < 7 but it will be mildly acidic medium, pH $\sim$ 3 - 5.

Reaction: $CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+$   [Note how dissociation of acetic acid is an equilibrium whereas HCl is fully dissociated]

Therefore, the tail of the curve (almost flat portion to the left) will be before the pH = 7, $\sim$ 3 - 5 .

[NOTE: for a weak base vs strong acid titration, e.g, NH3 vs HCl titration, before adding any NH3 only HCl will be present in the medium and hence solution will highly aicidic.]

ii. On adding NaOH: On addition of a few drops of NaOH, enough H3O+ ions are not present in the medium so the pH increase is fast for these initial 3-4 drops of NaOH.

But on further addition of NaOH, it will start to react with CH3COOH resulting in formation of substantial amount of salt, CH3COONa to create a buffer solution (the mixture of a weak acid and its salt with a strong base is a buffer solution.) So the change in pH is very low in this tail region of the curve.

Reaction occurring during titration: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

The pH for this region (before equivalence point) is determined by the Henderson-Hasselbach equation for buffers.

[NOTE: for a weak base vs strong acid titration, e.g, NH3 vs HCl titration, on addition of NH3 , HCl will be consumed and form NH4Cl salt. This a buffer solution and hence pH increase will be very slow and the tail will be flat in the highly acidic pH region

(pH = 1 - 3).]

2. At the equivalence point: All of the CH3COOH has been neutralized and hence only CH3COONa salt and water will be present in the medium. Now, since CH3COONa is a salt made from a weak acid and a strong base, it will undergo hydrolysis, and form NaOH in the process and hence the pH of the medium will be > 7. Therefore, the steep rise in the pH vs volume of NaOH added curve when equivalence has reached will be around the pH > 7 ( $\sim$ 8-9).

Hydrolysis Reaction: $CH_3COONa + H_2O \rightleftharpoons CH_3COOH + NaOH$

[NOTE: for a weak base vs strong acid titration, e.g, NH3 vs HCl titration, all of the HCl will be consumed and form NH4Cl salt which will undergo hydrolysis, produce a little amount of HCl and the pH will be in the acidic region.]

3. After the equivalence point: After all of the CH3COOH has been consumed by NaOH, addition of excess NaOH would result in excess of OH- ions and hence a basic medium, pH $\sim$ 12-13. Therefore, the head of the curve (almost flat portion to the right) will be after the pH = 7 , ( $\sim$ 12-13).

[NOTE: for a weak base vs strong acid titration, e.g, NH3 vs HCl titration, excess of NH3 (weak base) in the solution means weakly basic solution, in the pH range 8 - 9.]