# How to apply the Henderson–Hasselbalch equation when adding KOH to an acidic acid buffer?

Will the Henderson–Hasselbalch equation be able to calculate the pH when you have an acidic solution and you add a basic solution? Will that be possible or can you only use it for adding an acid to an acidic buffer to calculate the pH, or only a base to a basic buffer? If it can, how is the pH calculated, if you add 0.025 mol of solid KOH to 1 L of an acidic acid buffer solution of pH 3.97.

• In case of very dilute buffer solutions, you shold try the Brönsted–Charlot equation: en.wikipedia.org/wiki/Charlot_equation Aug 22 '17 at 8:34
• Asker123, when you originally asked this question (before people edited it), did you mean that we start with pure pH 3.97 acetic acid solution and add base in an attempt to create a buffer, or did you mean that we start with pH 3.97 buffer? Aug 24 '17 at 16:35

If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html

Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted.

pH = 14 - pOH

pH = 14 + log(0.025) = 12.4

In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example.

• You are adding OH- to the solution Feb 28 '15 at 17:21
• yes, because acetic acid concentration will be less than 0.001M, almost all the OH- will stay OH- Feb 28 '15 at 17:23
• Alright then but can you find the pH of a weak acid buffer when you add OH- with an equation?\ Feb 28 '15 at 18:24
• Yes, use the Henderson Hasselbach and don't add more OH- than weak acid Feb 28 '15 at 18:30
• The question makes no comment about the concentration of the buffer. It should however be noted, that pure acetic acid is no buffer, and your answer only accounts for this case. Aug 22 '17 at 12:50