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The Henderson-Hasselbalch Equation is: (1) a rearrangement of the K_{a} expression followed by (2) the use of negative logarithms. I will derive the H-H equation using the generic weak acid HA.

Here is the dissociation equation for HA:

HA ⇌ HFrom which, we write the K^{+}+ A¯

[H ^{+}] [A¯]K _{a}=–––––––– [HA]

Next, we isolate the H^{+} and put it on the left-hand side of the equation:

[HA] [H ^{+}] =K _{a}·––––– [A¯]

Now, I'm going to take the negative log of each of the three terms in the above equation. When I do that, they become:

1) −log [H^{+}]

2) −log K_{a}

3) −log ([HA] / [A¯])

However,

1) this is the pH

2) this is the pK_{a}

3) to get rid of the negative sign (don't ask why!), I simply flip the log term to get this: +log ([A¯] / [HA])

Inserting these last three items (the pH, the pK_{a} and the rearranged log term), we arrive at the Henderson-Hasselbalch Equation:

[A¯] pH = pK _{a}+ log––––– [HA]

Here is a common way the H-H equation is presented in a textbook explanation:

[salt form] pH = pK _{a}+ log––––––––– [acid form]

Remember that, in a buffer, the two substances differ by only a proton. The substance with the proton is the acid and the substance without the proton is the salt.

However, remember that the salt of a weak acid is a base (and the salt of a weak base is an acid).

Consequently, another common way to write the H-H equation is to substitute "base" for "salt form" (sometimes you will see "conjugate base" or "base form"). This is probably the most useful way to describe the interactions between the acidic form (the HA) and the basic form (the A¯).

Here it is:

[base] pH = pK _{a}+ log––––– [acid]

Remember this: the base is the one WITHOUT the proton and the acid is the one WITH the proton.

**An alternate form of the Henderson-Hasselbalch Equation:**

The above discussion used pH and pK_{a}. There is a alternate form of the HH equation using pOH and pK_{b}. It doesn't have a name and is seldom discussed. Having said that, let me issue a warning: your teacher will probably teach the HH equation and then could test the alternate form as well as the HH equation that was taught. Be warned!

Here is a discussion of the alternate form.

**Important comment**

Look at the log part of the H-H equation. Suppose that the salt form and the acid form were equal concentrations. That means that the ratio is 1 and the log of 1 is 0. The H-H equation then becomes:

pH = pK_{a}

This is an important (and useful) result. It can be used to determine an uknown K_{a} by carrying out a titration. The odds are quite good that you will be tested on this point and its usefulness. It is commonly discussed in the area of acid-base titrations.

**Last comment before moving on:**

In solving Henderson-Hasselbalch problems, it is very easy to make mistakes. The ChemTeam has made mistakes while teaching in the classroom as well as when writing the problems that follow. One common mistake is to switch the numbers that go in the numerator and denominator of the log term. Another is to transpose numbers.

Be careful out there. Lots of mistakes to be made!

Fifteen Buffer Examples | Buffer Problems 1-10 | Buffer Problems 11-20 | Buffer Problems 21-30 | Buffer Problems 31-40 |

Return to Introduction to Buffers | Return to the Acid Base menu |