Question
Question: How would you rearrange the Henderson-Hasselbalch equation to find out \([a/ha]\) from \(pH = pKa + ...
How would you rearrange the Henderson-Hasselbalch equation to find out [a/ha] from pH=pKa+log[a/ha]?
Solution
The Henderson-Hasselbalch equation is a mathematical equation which connects the pH of the solution and the pKa which is equal to the −logKa. The Ka is the acid dissociation constant of weak acid and its conjugate base.
Complete step by step answer:
The equation which relates the pH of an aqueous solution of an acid to the acid dissociation constant of the acid is described as the Henderson-Hasselbalch equation.
The equation is given as shown below.
pH=pKa+log([weakacid][Conjugatebase])
The pKa value is equal to the negative logarithm of acid dissociation constant of the weak acid. It measures the strength of the acid's solution. The weak acid has pKa value ranging from 2-12 in water.
It is given as shown below.
pKa=−log[Ka]
Where,
Ka is the acid dissociation constant of the weak acid.
The reaction of the weak acid-conjugate base buffer is shown below.
HA(aq)+H3O(l)⇄H3O+(aq)+A−(aq)
The pH of the solution is given as shown below.
pH=pKa+log([HA][A−])
Now, we need to determine the ratio which exists between the concentration of the conjugate base, A− and the concentration of the weak acid HA, add log on one side of the equation.
log([HA][A−])=pH−pKa
If x is equal to y,
10x=10y
The above equation is equivalent to
10log([HA][A−])=10pH−pKa
As we know,
10log10(x)=x
We get,
[HA][A−]=10pH−pKa
Note:
The Henderson-Hasselbalch equation is useful for determining the pH of the buffer solution and also determining the equilibrium pH in an acid-base reaction. The equation can also be used to determine the amount of acid and conjugate base used to prepare the buffer solution of a particular pH.