What is the function of buffer solution for titration with edta?
Answers
buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
What is a buffer composed of?
To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired pH when preparing the buffer. For example, the following could function as buffers when together in solution:
Acetic acid (weak organic acid w/ formula CH3COOH) and a salt containing its conjugate base, the acetate anion (CH3COO-), such as sodium acetate (CH3COONa)
Pyridine (weak base w/ formula C5H5N) and a salt containing its conjugate acid, the pyridinium cation (C5H5NH+), such as Pyridinium Chloride.
Ammonia (weak base w/ formula NH3) and a salt containing its conjugate acid, the ammonium cation, such as Ammonium Hydroxide (NH4OH)
How does a buffer work?
A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H3O+ and OH-) when the are added to the solution. To clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F- ion and solvated protons (H3O+), which does not allow it to dissociate completely in water. Therefore, if we obtain HF in an aqueous solution, we establish the following equilibrium with only slight dissociation (Ka(HF) = 6.6x10-4, strongly favors reactants):
HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)(1)(1)HF(aq)+H2O(l)⇌F(aq)−+H3O(aq)+
We can then add and dissolve sodium fluoride into the solution and mix the two until we reach the desired volume and pH at which we want to buffer. When Sodium Fluoride dissolves in water, the reaction goes to completion, thus we obtain:
NaF(aq)+H2O(l)→Na+(aq)+F−(aq)(2)(2)NaF(aq)+H2O(l)→Na(aq)++F(aq)−
Since Na+ is the conjugate of a strong base, it will have no effect on the pH or reactivity of the buffer. The addition of NaFNaF to the solution will, however, increase the concentration of F- in the buffer solution, and, consequently, by Le Châtelier’s Principle, lead to slightly less dissociation of the HF in the previous equilibrium, as well. The presence of significant amounts of both the conjugate acid, HFHF, and the conjugate base, F-, allows the solution to function as a buffer. This buffering action can be seen in the titration curve of a buffer solution.

As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted through neutralization. This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity.
If acid were added:
F−(aq)+H3O+(aq)⇌HF(aq)+H2O(l)(3)(3)F(aq)−+H3O(aq)+⇌HF(aq)+H2O(l)
In this reaction, the conjugate base, F-, will neutralize the added acid, H3O+, and this reaction goes to completion, because the reaction of F- with H3O+ has an equilibrium constant much greater than one. (In fact, the equilibrium constant the reaction as written is just the inverse of the Ka for HF: 1/Ka(HF) = 1/(6.6x10-4) = 1.5x10+3.) So long as there is more F- than H3O+, almost all of the H3O+ will be consumed and the equilibrium will shift to the right, slightly increasing the concentration of HF and slightly decreasing the concentration of F-, but resulting in hardly any change in the amount of H3O+ present once equilibrium is re-established.
If base were added:
HF(aq)+OH−(aq)⇌F−(aq)+