## The Henderson-Hasselbalch Equation

#### Learning Objective

• Calculate the pH of a buffer system using the Henderson-Hasselbalch equation.

#### Key Points

• The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction.
• The formula for the Henderson–Hasselbalch equation is: $pH=p{ K }_{ a }+log(\frac { { [A }^{ - }] }{ [HA] } )$, where pH is the concentration of [H+], pKa is the acid dissociation constant, and [A] and [HA] are concentrations of the conjugate base and starting acid.
• The equation can be used to determine the amount of acid and conjugate base needed to make a buffer solution of a certain pH.

#### Term

• pKaA quantitative measure of the strength of an acid in solution; a weak acid has a pKa value in the approximate range -2 to 12 in water and a strong acid has a pKa value of less than about -2.

The Henderson–Hasselbalch equation mathematically connects the measurable pH of a solution with the pKa (which is equal to -log Ka) 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 pKa 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 pKa, the solution of the Henderson-Hasselbalch equation gives the logarithm of a ratio which can be solved by performing the antilogarithm of pH/pK­a:

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

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+ (Ka 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