The Henderson-Hasselbalch equation is a mathematical expression that allows the calculation of the pH of a buffer or buffer solution. It is based on the pKa of the acid and the relationship between the concentrations of the conjugate base or salt and the acid, present in the buffer solution..
The equation was initially developed by Lawrence Joseph Henderson (1878-1942) in 1907. This chemist established the components of his equation based on carbonic acid as a buffer or buffer..
Later, Karl Albert Hasselbalch (1874-1962) introduced in 1917 the use of logarithms to complement the Henderson equation. The Danish chemist studied the reactions of blood with oxygen and the effect on its pH.
A buffer solution is able to minimize the pH changes that a solution undergoes by adding a volume of strong acid or base. It is made up of a weak acid and its strong conjugate base which dissociates quickly.
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A weak acid in an aqueous solution dissociates according to the Law of Mass Action, according to the following scheme:
HA + HtwoO ⇌ H+ + TO-
HA is the weak acid and A- its conjugate base.
This reaction is reversible and has an equilibrium constant (Ka):
Ka = [H+] ·[TO-] / [HA]
Taking logarithms:
log Ka = log [H+] + Log [A-] - log [HA]
If each term of the equation is multiplied by (-1), it is expressed as follows:
- log Ka = - log [H+] - log [A] + log [HA]
The - log Ka is defined as pKa and the - log [H+] is defined as pH. After making the proper substitution, the mathematical expression reduces to:
pKa = pH - log [A-] + Log [HA]
Solving for pH and regrouping terms, the equation is expressed as follows:
pH = pKa + log [A-] / [HA]
This is the Henderson-Hasselbalch equation for a weak acid buffer.
Similarly, a weak base can form a buffer, and the Henderson-Hasselbalch equation for it is as follows:
pOH = pKb + log [HB] / [B-]
However, most of the buffers are originated, even those of physiological importance, from the dissociation of a weak acid. Therefore, the most used expression for the Henderson-Hasselbalch equation is:
pH = pKa + log [A-] / [HA]
The Henderson-Hasselbalch equation indicates that this solution is made up of a weak acid and a strong conjugate base expressed as a salt. This composition allows the buffer solution to remain at a stable pH even when strong acids or bases are added..
When strong acid is added to the buffer, it reacts with the conjugate base to form a salt and water. This neutralizes the acid and allows the pH variation to be minimal..
Now, if a strong base is added to the buffer, it reacts with the weak acid and forms water and a salt, neutralizing the action of the added base on the pH. Therefore, the pH variation is minimal.
The pH of a buffer solution depends on the ratio of the concentrations of the conjugate base and the weak acid, and not on the absolute value of the concentrations of these components. A buffer can be diluted with water and the pH will remain virtually unchanged.
The buffering capacity also depends on the pKa of the weak acid, as well as the concentrations of the weak acid and the conjugate base. The closer the pH of the buffer is to the pKa of the acid, the greater its buffering capacity..
Also, the higher the concentration of the components of the buffer solution, the greater its buffering capacity..
pH = pKa + log [CH3COO-] / [CH3COOH]
pKa = 4.75
pH = pKa + log [HCO3-] / [HtwoCO3]
pKa = 6.11
However, the overall process that leads to the formation of the bicarbonate ion in a living organism is as follows:
COtwo + HtwoO ⇌ HCO3- + H+
Being the COtwo a gas, its concentration in solution is expressed as a function of its partial pressure.
pH = pka + log [HCO3-] / α pCOtwo
α = 0.03 (mmol / L) / mmHg
pCOtwo is the partial pressure of COtwo
And then the equation would look like:
pH = pKa + log [HCO3-] / 0.03 pCOtwo
pH = pKa + log [lactate ion] / [lactic acid]
pKa = 3.86
pH = pKa + log [dibasic phosphate] / [monobasic phosphate]
pH = pKa + log [HPO4two-] / [HtwoPO4-]
pKa = 6.8
pH = pKa + log [HbOtwo-] / [HHbOtwo]
pKa = 6.62
pH = pKa + log [Hb-] / HbH
pKa = 8.18
The phosphate buffer is important in regulating body pH, since its pKa (6.8) is close to the existing pH in the body (7.4). What will be the value of the relation [NatwoHPO4two-] / [NaHtwoPO4-] of the Henderson-Hasselbalch equation for a pH value = 7.35 and a pKa = 6.8?
The NaH dissociation reactiontwoPO4- it is:
NaHtwoPO4- (acid) ⇌ NaHPO4two- (base) + H+
pH = pKa + log [NatwoHPO4two-] / [NaHtwoPO4-]
Solving for the [conjugate base / acid] ratio for the phosphate buffer, we have:
7.35 - 6.8 = log [NatwoHPO4two-] / [NaHtwoPO4-]
0.535 = log [NatwoHPO4two-] / [NaHtwoPO4-]
100.535 = 10log [Na2HPO4] / [NaH2PO4]
3.43 = [NatwoHPO4two-] / [NaHtwoPO4-]
An acetate buffer has an acetic acid concentration of 0.0135 M and a sodium acetate concentration of 0.0260 M. Calculate the pH of the buffer, knowing that the pKa for the acetate buffer is 4.75.
The dissociation equilibrium for acetic acid is:
CH3COOH ⇌ CH3COO- + H+
pH = pKa + log [CH3COO-] / [CH3COOH]
Substituting the values we have:
[CH3COO-] / [CH3COOH] = 0.0260 M / 0.0135 M
[CH3COO-] / [CH3COOH] = 1.884
log 1.884 = 0.275
pH = 4.75 + 0.275
pH = 5.025
An acetate buffer contains 0.1 M acetic acid and 0.1 M sodium acetate. Calculate the pH of the buffer after adding 5 mL of 0.05 M hydrochloric acid to 10 mL of the above solution.
The first step is to calculate the final concentration of the HCl when mixed with the buffer solution:
ViCi = VfCf
Cf = Vi · (Ci / Vf)
= 5 mL · (0.05 M / 15 mL)
= 0.017 M
Hydrochloric acid reacts with sodium acetate to form acetic acid. Therefore, the sodium acetate concentration decreases by 0.017 M and the acetic acid concentration increases by the same amount:
pH = pKa + log (0.1 M - 0.017 M) / (0.1 M + 0.017 M)
pH = pKa + log 0.083 / 0.017
= 4.75 - 0.149
= 4.601
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