Henderson equation

Henderson and Hasselbalch equations can be utilised to evaluate the pH of a buffer solution. The mathematical value of the acid separation constant, Ka of the acid, is predicted or known.

The pH is measured for a specified focus value of acid, HA, and sodium chloride, MA of the associated base, A–. For instance, the compound may contain sodium acetate and acetic acid.

A Henderson and Hasselbalch equation that can measure the value of pH based on the given buffer compound was developed by Lawrence Joseph Henderson in 1908, an American chemist. Afterwards, Karl Albert Hasselbalch reformed this equation from a logarithmic perspective. Henderson equation defines a bond between the pH of acids (melted reaction) and their pKa (acid separation constant).

The pH buffer reaction can be derived with this equation when the focus of the acid and joint base, or the ground and the respective common acid, are known.

Following is the Henderson equation to evaluate pH value depending upon the provided buffers solution.

pH = pKa + log10 ( Base/ Acid)

Where pH refers to the acidity of a buffer reaction,

pKa represents Ka’s negative logarithm

Ka. means acid separation constant

Base refers to a focus of a joint base

HA refers to a principle of an acid

Deriving Henderson and Hasselbalch Equation

Some straightforward methods can measure the persistence of ionisation of solid acids and strong bases but might not be utilised with low acids and bases. As a proposition of ionisation of these acids and bases are extremely weak. Hence, to measure the pH of these solutions, the Henderson Hasselbalch equation comes into the picture.

For instance, ionisation of a low acid HA:

HA + H2O ⇋ H+ + A– 

Acid separation constant, Ka can be denoted as,

Ka = [H+][A–]/[HA]

Using negative logarithm of RHS and LHS:

-log Ka = -log  [H+][A–]/[HA]

-log Ka = -log  [H+] – log [A–]/[HA]

Because -log[H+] = pH and -log Ka = pKa,

Above stated expression can also be written as,

pKa = pH – log [A–]/[HA]

Repositioning the equation,

pH = pKa + log [A–]/[HA] – equation 1

The equation 1 is the Henderson Hasselbalch equation, commonly pronounced as the Henderson equation. It is a fundamental equation to measure the equilibrium pH in acid-base compounds. Through the equation, we can conclude that when pH = pKa

log [A–]/[HA] = 0

[A–] = [HA]

When PH = pKa, the focus of both strains is similar, or we can say the acid will be half separated.

In the same way for low base B:

B + H2O ⇋ OH– + HB+

Base separation constant, Kb, of the base mentioned as,

Kb = [BH+][OH–]/[B]

Using a negative log of RHS and LHS,

-logKb = -log [BH+][OH–]/[B]

-logKb = -log [OH–] – log [BH+]/[B]

Because -log [OH–] = pOH and -logKb = pKb,

Expression stated above can also be written as,

pKb = pOH – log[BH+]/[B]

Repositioning the equation,

pOH = pKb + log [BH+]/[B]

Following are statements you need to ensure while using the equation:

• When half of the acid goes through separation, then the value of [A]/[HA]  tends to be 1, assuming that the pKa acid is exact to the pH of the reaction at this time. (pH = pKa + log10(1) = pKa).
• For each single unit alteration in the pH to pKa proposition, a tenfold change happens in the proportion of the related acid to the separated acid. 
• For instance, when the pKa acid value is seven and the pH of the reaction is 6, the value of [A–]/[HA] becomes 0.1. However, if the pH of the reaction is 5, then the value of [A–]/[HA] will be 0.01.
• The value of [A–]/[HA] is derived based on the value of pH and pKa. When pH < pKa then [A–]/[HA]  < 1 and if pH > pKa; then [A–]/[HA] > 1.

Henderson Hasselbalch Equation Example

Calculate the pH value of a buffer reaction retrieved from 0.20 M CH3COOH and 0.50 M CH3COO–, which has an acid separation constant for CH3COOH of 1.8 x 10-5.

Solving this complication by applying the value of Henderson equation for a weak acid along with joint base.

Henderson Hasselbalch equation : pH = pKa + log ([A-]/[HA]),

Apply values to the equation: pH = pKa + log ([CH3COO–] / [CH3COOH]) then,

pH = -log (1.8 x 10-5) + log (0.50 M / 0.20 M)

pH = -log (1.8 x 10-5) + log (2.5)

pH = 4.7 + 0.40

pH = 5.1 

Hence, post applying the values to the equation, we get the pH value of the solution as 5.1.

In such a way, we can retrieve the pH value by referring to the Henderson Hasselbalch equation example. Apply the given discount into the equation, and you are good to go.

Conclusion

The Henderson equation is a straightforward buffer solution containing an acid and sodium chloride of the joint base of the acid. The Henderson Hasselbalch equation associates the pH of a solution consisting of a compound of two elements to the acid separation constant, Ka, and focus of the strain in solution.

Henceforth knowing pH, pKa, buffer solutions, and the Henderson Hasselbalch equation is extremely important for better understanding while resolving such problems.

We’ve covered some essential points that you need to know while Deriving Henderson and Hasselbalch Equation. Using the equation, you can also refer to the Henderson Hasselbalch example and solve other problems.