What is the Henderson-Hasselbalch equation?

Answer 1

Note- See the related link for the complete derivation below

SUMMARY OF ACID-DISSOCIATION CONSTANT (pKa) (from Rhoades and Pflanzer Human Physiology)

HA ßà H+ + A-

1) Reaction to the right à dissociation reaction

2) Reaction to left à association reaction

The rate of the dissociation reaction = [HA] x dissociation rate constant k1 (which is a specific value for this reaction).

The rate of the association reaction = [H+] x [A-] x association rate constant k2

At equilibrium à rates of association and dissociation are =. Therefore

k1 x [HA] = k2 x [H+] x [A-]

Hence à [H+] x [A-] /[HA] = k1/k2

A NEW CONSTANT à is defined for k1/k2 à we call it Ka (equilibrium constant for the reaction and dissociation constant for the acid)

A HIGHER Ka à more completely an acid is dissociated à stronger acid à lower pH

A LOWER Ka à not as much dissociation à weak acid à higher pH

The Ka is often small in difficult to manipulate à so we present the number in a logarithmic form à pKa (which is the log10 of the INVERSE of Ka

pKa = log10(1/Ka) = --log10(Ka)

LOW pKaà high dissociation constant à STRONG ACID

HIGH pKa à low dissociation constant à WEAK ACID

THE HENDERSON HASSELBALCH EQUATION

[H+] x [A-] /[HA] = Ka à therefore

[H+] = Ka x [HA] / [A-]

Take log of both side

log[H+] = logKa + log([HA]/[A-]) à multiple both sides by -1

-- log[H+] = --logKa + log([A-]/[HA])

And because pH = --log[H+] and pKa = log(1/Ka) = --log(Ka)

pH = pKa + log([A-]/[HA])

HENCE à WHEN [A-] = [HA] à the pH of solution = it's pKa (because the log1 is 0)

Conversely à the pKa is the pH at which there are as many molecules of weak acid as there are conjugate base in solution.

For the bicarbonate buffer system à (pK = 6.1)

Cheers

Answer 2

Henderson heselbach equation is depends on the pKa of the weak acid and the ratio of molar concentration of a salt and weak acid.

Answer 3

Its an equation you can use to find the pH of a solution.

it is.... --- pH = pKa + log (Base/Acid) ---

these may help too

Ka = 10^-pKa

Kw = Ka*Kb