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Preparing Buffer Solutions

When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.

Where did the Henderson-Hasselbalch Equation come from?

Where the Henderson-Hasselbalch approximation comes from:

\(HA + H_2O \rightleftharpoons H_3O^+ + A^-\)


\(A^-\) = conjugate base

\(HA\) = weak acid 

We know that \(K_a\) is equal to the products over the reactants and, by definition, H2O is essentially a pure liquid that we consider to be equal to one.

\(K_a = [H_3O^+][A^-]\)

Take the -log of both sides:

\(-log \; K_a = -log([H_3O^+][A^-])\)

\(-log \; K_a = -log[H_3O^+] \; -log[A^-]\)

Using the following two relationships:

\(-log[K_a] = pK_a\)

\(-log[H_3O^+] = pH\)

We can simplify the above equation:

\(pK_a = pH - log[A^-]\)

If we add log[A-] to both sides, we get the Henderson-Hasselbalch Equation: 

\(pH = pK_a + log[A^-]\)

This equation is only valid when…

  1. The conjugate base / acid falls between the values of 0.1 and 10
  2. The molarity of the buffers exeeds the value of the Ka by a factor of at least 100

Preparing Buffer Solutions

There are two cases where we can use the Henderson-Hasselbalch Equation.

Case #1

Suppose we needed to make a buffer solution with a pH of 2.11. In the first case, we would try and find a weak acid with a pKa value of 2.11


at the same time the molarities of the acid and the its salt must be equal to one another. This will cause the two molarities to cancel; leaving the log [A-]  equal to log(1) which is zero.

\(pH = pK_a + log[A^-] = 2.11 + log(1) = 2.11\)

This is a very unlikely scenario, however, and you won't often find yourself with Case #1

Case #2

An Example:

What mass of NaC7H502 must be dissolved in 0.200 L of 0.30 M HC7H5O2 to produce a solution with pH = 4.78? (Assume solution volume is constant at 0.200L)


\(HC_7H_5O_2 + H_20 \rightleftharpoons H_3O^+ + C_7H_5O_2\)

\(K_a =6.3 \times 10^{-5}\)

\(K_a = \dfrac{[H_3O^+][C_7H_5O_2]}{[HC_7H_5O_2]} = 6.3 \times 10^{-5}\) 

\([H_3O^+] = 10^{-pH} = 10^{-4.78} = 16.6 \times 10^{-6}\;M\;[HC_7H_5O_2] = 0.30\;M\;[C_7H_5O_2] =\)

\([C_7H_5O_2^-] = K_a \times \dfrac{[HC_7H_5O_2]}{[H_3O^+]}\)

\(1.14 \; M = 6.3 \times 10^{-5} \times \dfrac{0.30}{16.6 \times 10^{-6}}\)

Mass = 0.200 L x 1.14 mol C7H5O2- / 1L x 1mol NaC7H5O2 / 1 mol C7H5O2- x 144 g NaC7H5O2 / 1 mol NaC7H5O2 = 32.832 g NaC7H5O2


  1. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007.


  • Jonathan Nguyen (UCD), Garrett Larimer (UCD)

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Last Modified
09:23, 2 Oct 2013

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