The Henderson-Hasselbalch Equation

The Henderson–Hasselbalch equation mathematically connects the measurable pH of a solution with the pK_a (which is equal to -log K_a) of the acid. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction. The equation can be derived from the formula of pK_a for a weak acid or buffer. The balanced equation for an acid dissociation is:

$HA\rightleftharpoons { H }^{ + }+{ A }^{ - }$

The acid dissociation constant is:

${ K }_{ a }=\frac { [{ H }^{ + }][A^{ - }] }{ [HA] }$

After taking the log of the entire equation and rearranging it, the result is:

$log({ K }_{ a })=log[{ H }^{ + }]+log(\frac { { [A }^{ - }] }{ [HA] } )$

This equation can be rewritten as:

$-p{ K }_{ a }=-pH+log(\frac { [A^{ - }] }{ [HA] } )$

Distributing the negative sign gives the final version of the Henderson-Hasselbalch equation:

$pH=p{ K }_{ a }+log(\frac { { [A }^{ - }] }{ [HA] } )$

In an alternate application, the equation can be used to determine the amount of acid and conjugate base needed to make a buffer of a certain pH. With a given pH and known pK_a, the solution of the Henderson-Hasselbalch equation gives the logarithm of a ratio which can be solved by performing the antilogarithm of pH/pK­_a:

${ 10 }^{ pH-p{ K }_{ a } }=\frac { [base] }{ [acid] }$

An example of how to use the Henderson-Hasselbalch equation to solve for the pH of a buffer solution is as follows:

What is the pH of a buffer solution consisting of 0.0350 M NH₃ and 0.0500 M NH₄⁺ (K_a for NH₄⁺ is 5.6 x 10^-10)? The equation for the reaction is:

${NH_4^+}\rightleftharpoons { H }^{ + }+{ NH_3}$

Assuming that the change in concentrations is negligible in order for the system to reach equilibrium, the Henderson-Hasselbalch equation will be:

$pH=p{ K }_{ a }+log(\frac { { [NH_3}] }{ [NH_4^+] } )$

$pH=9.25+log(\frac{0.0350}{0.0500} )$

pH = 9.095