The buffer solutions or buffers are those that can decrease pH changes due to H ions3OR+ and OH-. In the absence of these, some systems (such as physiological ones) are harmed, as their components are very sensitive to sudden changes in pH.
Just as shock absorbers in cars reduce the impact caused by their movement, buffers do the same but with the acidity or basicity of the solution. Furthermore, buffers establish a specific pH range within which they are efficient..
Otherwise, the H ions3OR+ They will acidify the solution (the pH drops to values below 6), resulting in a possible alteration in the reaction yield. The same example can be applied for basic pH values, that is, greater than 7.
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They are essentially composed of an acid (HA) or a weak base (B), and salts of their conjugated base or acid. Consequently, there are two types: acid buffers and alkaline buffers..
Acid buffers correspond to the HA / A pair-, where to- is the conjugate base of the weak acid HA and interacts with ions - such as Na+- to form sodium salts. Being this way, the pair remains as HA / NaA, although they can also be potassium or calcium salts.
Derived from the weak acid HA, it buffers acidic pH ranges (less than 7) according to the following equation:
HA + OH- => A- + HtwoOR
However, being a weak acid, its conjugate base is partially hydrolyzed to regenerate part of the HA consumed:
TO- + HtwoOR <=> HA + OH-
On the other hand, alkaline buffers consist of the B / HB pair+, where HB+ it is the conjugated acid of the weak base. Generally, HB+ forms salts with chloride ions, leaving the pair as B / HBCl. These buffers buffer basic pH ranges (greater than 7):
B + H3OR+ => HB+ + HtwoOR
And again, HB+ can be partially hydrolyzed to regenerate part of consumed B:
HB+ + HtwoOR <=> B + H3OR+
While acidic buffers buffer acidic pH and alkaline buffers basic pH, both can react with H ions.3OR+ and OH- through these series of chemical equations:
TO- + H3OR+ => HA + HtwoOR
HB+ + Oh- => B + HtwoOR
Thus, in the case of the HA / A pair-, HA reacts with OH ions-, while A- -its conjugate base- reacts with the H3OR+. Regarding the B / HB pair+, B reacts with H ions3OR+, while HB+ -its conjugated acid- with the OH-.
This allows both buffers to neutralize both acidic and basic species. The result of the above versus, for example, the constant addition of moles of OH-, is the decrease in the pH variation (ΔpH):
The upper image shows the buffering of the pH against a strong base (donor of OH-).
Initially the pH is acidic due to the presence of HA. When the strong base is added, the first moles of A are formed- and the buffer starts to take effect.
However, there is an area of the curve where the slope is less steep; that is, where the damping is more efficient (bluish box).
There are several ways to understand the concept of damping efficiency. One of these is to determine the second derivative of the curve pH versus the volume of base, solving for V for the minimum value, which is Veq / 2.
Veq is the volume at the equivalence point; this is the volume of base needed to neutralize all the acid.
Another way to understand it is through the famous Henderson-Hasselbalch equation:
pH = pKto + log ([B] / [A])
Here B denotes the base, A the acid, and pKto is the smallest logarithm of the acidity constant. This equation applies to both the acid species HA and the conjugated acid HB+.
If [A] is very large with respect to [B], the log () takes a very negative value, which is subtracted from the pKto. If, on the other hand, [A] is very small with respect to [B], the value of log () takes on a very positive value, which is added to pKto. However, when [A] = [B], the log () is 0 and the pH = pKto.
What does all of the above mean? That the ΔpH will be greater in the extremes considered for the equation, while it will be minimum with a pH equal to the pKto; and as the pKto is characteristic of each acid, this value determines the pK rangeto± 1.
The pH values within this range are those in which the buffer is most efficient..
To prepare a buffer solution, you need to keep the following steps in mind:
- Know the required pH and, therefore, the one you want to keep as constant as possible during the reaction or process.
- Knowing the pH, one looks for all weak acids, those whose pKto is closer to this value.
- Once the HA species has been chosen and the concentration of the buffer calculated (depending on how much base or acid needs to be neutralized), the necessary amount of its sodium salt is weighed..
Acetic acid has a pKto of 4.75, CH3COOH; therefore, a mixture of certain amounts of this acid and sodium acetate, CH3COONa, form a buffer that efficiently buffers in the pH range (3.75-5.75).
Other examples of monoprotic acids are benzoic acids (C6H5COOH) and formic (HCOOH). For each of these their pK valuesto they are 4.18 and 3.68; therefore, their highest buffering pH ranges are (3.18-5.18) and (2.68-4.68).
On the other hand, polyprotic acids such as phosphoric (H3PO4) and carbonic (HtwoCO3) have as many pK valuesto as protons can release. Thus, the H3PO4 has three pKto (2.12, 7.21 and 12.67) and the HtwoCO3 has two (6,352 and 10,329).
If you want to maintain a pH of 3 in a solution, you can choose between HCOONa / HCOOH (pKto= 3.68) and NaHtwoPO4/ H3PO4 (pKto= 2.12).
The first buffer, that of formic acid, is closer to pH 3 than the phosphoric acid buffer; therefore HCOONa / HCOOH buffers better at pH 3 than NaHtwoPO4/ H3PO4.
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