In 1908, Lawrence Joseph Henderson derived an equation to calculate the \([H_3O^{+}]\) in a buffer solution. In 1909, Søren Peter Lauritz Sørenson introduced the idea of pH, which allowed Karl Albert Hasselbalch to rework Henderson’s equation resulting in what is known today as the Henderson-Hasselbalch equation:
\(pH = pK_a - log(\frac{[HA]}{[A^{-1}]})\)
Where \(pK_a\) is the -log of the acid dissociation constant (\(K_a\)), \([HA]\) is teh molar concentration of a weak acid, and \([A^{-1}]\) is the molar concentration of the conjugate base.
The equation is also written like this:
\(pH = pK_a + log(\frac{[A^{-1}]}{[HA]})\)
This is mathematically identical.
The Derivation
Although it is unlikely that you would need to know the derivation of the equation, seeing the derivation may help you understand when the equation can be used and its limitations.
We begin with the acid dissociation reaction and the equilibrium expression for that reaction:
\(HA + H_2O \rightleftharpoons A^{-1} + H_3O^{+}~~~~~~~~~~~K_a = \frac{[A^{-1}][H_3O^{+}]}{[HA]}\)
We can algebraically rearrange the equilibrium expression to solve for \([H_3O^{+}\) giving this equation:
\([H_3O^{+}] = K_a \cdot \frac{[HA]}{[A^{-1}]}\)
Taking the negative log of everything, and remembering that \(log(x \cdot y) = log(x) + log(y) gives us:
\(-log[H_3O^{+}] = -log(K_a) - log(\frac{[HA]}{[A^{-1}]})\)
which translates into:
\(pH = pK_a - log(\frac{[HA]}{[A^{-1}]})\)
Understanding and using the Henderson-Hasselbalch Equation
Like any good equation, the Henderson-Hasselbalch equation is only useful if you KNOW all but one of the variables. In other words, you can only use this equation to solve for pH if you KNOW the \(K_a\) and both the \([H_3O^+]\) and the \([OH^{-1}]\).
The good news for us, is that weak acids and bases have small \(K_a\) and \(K_b\) values which leads to “little x” problems. Take a look at this ice table:
Assuming that both “some” and “a bit” are non-zero amounts, then if this is a “little x” problem, those concentrations won’t (appreciably) change. So, in that situation, we already “know” the \([HA]\) and the \([A^{-1}]\).
The upside is that when you have a non-zero amount of both weak acid and conjugate base present in the solution, the Henderson-Hasselbalch equation can be used to find the pH.
Buffer solutions
We can use the Henderson-Hasselbalch equation to find the pH of a buffer solution when given an amount of both a weak acid and its conjugate base. For instance:
What is the pH of a buffer solution that has a \([HC_2H_3O_2]\) = 0.25 M and a \([C_2H_3O_2^{-1}]\) = 0.20 M. The \(K_a\) of acetic acid is \(1.8x10^{-5}\).
The solution:
\(pH = -log(1.8x10^{-5}) - log(\frac{0.25 M}{0.20 M}) = 4.65\)
Finding the pH during a weak titration
During a weak titration (for instance, a weak acid being titrated by a strong base), there is a region within which the Henderson-Hasselbalch equation can be used. This region falls between roughly 10% of the equivalence point volume and 90% of the equivalence point. This region is called the buffer region. These restrictions ensure that the “some” and “a bit” in the ICE table above are large enough so that the “little x” can be ignored. Let’s look at the 10.00 mL point in the problem below. This problem is FULLY worked out here.
50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?
The 10.00 mL point corresponds to this:
So, the pH can be found by doing this:
\(pH = -log(2.56x10^{-5}) - log(\frac{(\frac{0.09 mols}{0.060 L})}{(\frac{0.010 moles}{0.060 L})}) = 3.64\)
Of course, since the acid and base are in the same solution and, therefore, have the same volume, you could also just have done this:
\(pH = -log(2.56x10^{-5}) - log(\frac{(0.09 mols)}{(0.010 moles)}) = 3.64\)
Finding \(K_a\) by weak titration
At the beginning of a weak titration the solution contains “only” the weak acid, HA. At the equivalence point, all of those acid molecules have lost their \(H^+\) and are now \(A^{-1}\). So, logically, when you are halfway to the equivalence point, then half of the acid molecules will remain and half will have been converted to \(A^{-1}\).
That means that halfway to the equivalence point the ratio \(\frac{[HA]}{[A^{-1}]}\) = 1. Since the -log(1) = 0, then \(pH = pK_a\). That point is called the maximum buffer point and we can use that idea to solve a problem like this one:
A student titrates 100.0 mL of 0.100 M HA with 0.200 M NaOH. After 25.00 mL of the base have been added, the pH of the solution is 6.12. What is the value of \(K_a\)?
The equivalence point of the titration is 50.00 mL of NaOH, so 25.00 mL is halfway. At that point \(pH = pK_a\). So,
\(10^{pH} = 10^{pK_a} ~~~~~~~~~ K_a = 7.59x10^{-7}\)
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