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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Acids and Bases > Buffers > Henderson-Hasselbalch Approximation

Henderson-Hasselbalch Approximation

The Henderson-Hasselbalch formula allows us one method to approximate the pH of a buffer solution. The basic equation is as follows:

\[ pH = pK_a + log\dfrac{[A^-]}{[HA]} \]

Motivation

We have straightforward caluclations for strong acids and bases, but the computations behind buffers are rather complex and time consuming. By using the fact that weak acids and bases barely ionize, allowing us to approximate the pH of buffer solutions using initial concentrations. Though the approximation has a few restrictions, it simplifies a lenghty calculation into a simple equation derived from K.

Brief History

  • Lawrence Joseph Henderson (1878-1942) was a talented biochemist, among many other titles, who spent most of his career at Harvard. He was responsible for developing the components of the equation after studying equilibrium reactions that took place within blood as a result of respiration (specializing in "fatigue"). His equation was incomplete without a solid calculations going into it.M
  • Karl Albert Hasselbalch (1874-1962) was a chemist who studied pH closely. He also studied blood and reactions that took place with oxygen, to put in the simplest of terms. He eventually modified Henderson's equation by putting mathematical logs into it creating a solid relationship.J

Requirements

  1. \[-1 < \log \dfrac{[A^-]}{[H\!A]}  < 1\]
  2. The molarity of the buffer(s) should be 100x that of the acid ionization constant, or Ka.
  3. This equation will give poor or inaccurate results if there are strong acids or bases. pKa values between 5 and 9 will give good approximations, but when we are out of this range there is a strong chance that the pH value will be incorrect.J  In other words near the end or the beginning of a titration where the "relative concentrations of the acid or base differ substantially[,]... approximate calculations break down."J

Derivation

For a weak acid \( HA \) and its conjugate base \(A^-\):

\(H\!A + H_2O\leftrightarrows H^+ + A^-\)

which has an acid ionization constant

\[K_a=\frac{[H^+][A^-]}{[H\!A]}\]

The Henderson-Hasselbalch equation is derived from this acid ionization constant.

\[K_a= \frac{[H^+][A^-]}{[H\!A]}\]

\[-logK_a= -\log\frac{[H^+][A^-]}{[H\!A]}\]

\[- logK_a = - \log[H^+] - \log\frac{[A^-]}{[H\!A]}\]

\[pK_a = pH - \log\frac{[A^-]}{[H\!A]}\]

\[pH = pK_a + \log\frac{[A^-]}{[H\!A]}\]

Note that \( H\!A\)is a weak acid.  By definition, \( "H\!A\) does not dissociate completely and we can say  \( [H\!A]\approx[H\!A]_i\) and \( [A^-]\approx[A^-]_i\). Hence, we can use the initial concentrations because

\[pK_a + \log\frac{[A^-]_i}{[H\!A]_i} \approx pK_a + \log\frac{[A^-]}{[H\!A]} = pH\]

Similarly, for a weak base \( B\) and its conjugate acid \(H\!B^+\):

\[B + H_2O\leftrightarrows O\!H^- + H\!B^+\]

which has an base ionization constant

\[K_b=\frac{[O\!H^-][H\!B^+]}{[B]}\]

The Henderson-Hasselbalch equation is derived from this base ionization constant.

\[K_b= \frac{[O\!H^-][H\!B^+]}{[B]}\]

\[-logK_b= -\log\frac{[O\!H^-][H\!B^+]}{[B]}\]

\[- logK_b = - \log[O\!H^-] - \log\frac{[H\!B^+]}{[B]}\]

\[pK_b = pO\!H - \log\frac{[H\!B^+]}{[B]}\]

\[pO\!H = pK_b + \log\frac{[H\!B^+]}{[B]}\]

Note that \( B\) is a weak base.  By definition, \( "B\) does not dissociate completely and we can say  \( [B] \approx [B]_i\) and \( [H\!B^+] \approx [H\!B^+]_i\). Hence, we can use the initial concentrations because

\[pK_b + \log\frac{[H\!B^+]_i}{[B]_i} \approx pK_b + \log\frac{[H\!B^+]}{[B]} = pOH\]

Problems

  1. Find [H+] in a solution 1.0M HNO2 and 0.225M NaNO2 (Ka NO2 = 7.4*10-4)
  2. What ratio \( \frac{[A^-]}{[HA]} \) will create an acetic acid buffer of pH 5.0? (Ka acetic acid = 1.8*10-5)
  3. You prepare a buffer solution of 0.323 M NH3 and (NH4)2SO4. What molarity of (NH4)2SO4 is necessary to have a pH of 8.6? (pKbNH3= 4.74)
  4. What is the pH of a buffer 0.500 moles acetic acid and 0.500 moles acetate ion and the total volume is 5 L when you add 0.350 moles HCl? (Ka acetic acid = 1.8*10-5)
  5.  What is the range of an acetic acid buffer described in (4.) without the added HCl?

Solutions

(1.) pKa = -log Ka = -log(7.4*10-4) = 3.14
pH = pKa + log (\( \frac{[NO_2^-]}{[HNO_2]}\))
pH = 3.14 + log ( 1/0.225 )
pH = 3.14 + 0.648 = 3.788
[H+] = 10-pH = 10-3.788 = 1.6*10-4

(2.) pKa = -log Ka = -log(1.8*10-5) = 4.74
pH = pKa + log (\( \frac{[A^-]}{[H\!A]}\))
5.0 = 4.74 + log (\( \frac{[A^-]}{[H\!A]}\))
0.26 = log (\( \frac{[A^-]}{[H\!A]}\))
100.26 = \( \frac{[A^-]}{[H\!A]}\)
1.8 = \( \frac{[A^-]}{[H\!A]}\)

(3.) pKa+ pKb = 14
pKa= 14 - 4.74 = 9.26
pH = pKa + log (\( \frac{[A^-]}{[H\!A]}\))
8.6 = 9.26 + log (\( \frac{0.323}{[N\!H_4^+]}\))
-0.66 = log (\( \frac{0.323}{[N\!H_4^+]}\))
[NH4+] = 1.48 M

(4.) pKa = -log Ka = -log(7.4*10-4) = 3.14
pH = pKa + log (\( \frac{[acetate]}{[acetic acid] + [HCL]}\)) (note that mole ratio also works in place of concentrations as both the acid and base are in the same solution)
pH = 3.14 + log (0.588)
pH = 3.14 - .230 = 3.37

(5.) pKa = -log Ka = -log(7.4*10-4) = 3.14
Note that the buffer range is of magnitude 2 pH units, with the pKaas the midpoint.
Hence, the buffer range would be 2.14 to 4.14.

References

  1. J"Henderson-Hasselbalch Equation: Its History and Limitations," Henry N. Po and N. M. Senozan, J. Chem. Educ., 2001, 78 (11), p 1499
  2. M National Academy of Sciences (U.S.) Biographical memoirs. City of Washington, 1945. Vol. XXIII, 2d memoir.
  3. Chem. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007.

Contributors

  • Gurinder Khaira, Alexander Kot (UCD)

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Last Modified
14:14, 25 Feb 2014

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