Buffer pH Calculator (Henderson-Hasselbalch)

About the Buffer pH Calculator (Henderson-Hasselbalch)

The Henderson-Hasselbalch equation gives the pH of a buffer from the acid's pKa and the ratio of conjugate base [A⁻] to weak acid [HA] concentrations.

How to use this calculator

1. Enter the \(pK_a\) value for the weak acid in the buffer.
2. Enter the concentration or amount ratio of conjugate base to acid, \(\frac{[A^-]}{[HA]}\).
3. Read the calculated buffer pH from the result.
4. If needed, adjust the ratio to see how the pH changes when the buffer becomes more basic or more acidic.

The formula explained

The formula \(pH = pK_a + \log_{10}\!\left(\frac{[A^-]}{[HA]}\right)\) estimates the pH of a buffer solution. It compares the amount of conjugate base \([A^-]\) to the amount of weak acid \([HA]\), so a larger ratio gives a higher pH and a smaller ratio gives a lower pH.

• pH = the estimated acidity of the buffer solution
• \(pK_a\) = the negative logarithm of the acid dissociation constant for the weak acid
• \([A^-]\) = the concentration or amount of conjugate base
• [HA] = the concentration or amount of weak acid
• \(\frac{[A^-]}{[HA]}\) = the ratio of conjugate base to weak acid

Step by step method

1. Identify the weak acid in the buffer and note its \(pK_a\).
2. Find the ratio \(\frac{[A^-]}{[HA]}\) using concentrations or amounts in the same units.
3. Take \(\log_{10}\) of that ratio, then add the result to \(pK_a\).
4. Check whether the pH makes sense, a ratio greater than \(1\) makes the solution more basic, while a ratio less than \(1\) makes it more acidic.

Worked example

Problem. A buffer has \(pK_a = 4.76\), with \([A^-] = 0.20\,\text{M}\) and \([HA] = 0.10\,\text{M}\). Find the buffer pH.

1. Compute the ratio, \(\frac{[A^-]}{[HA]} = \frac{0.20}{0.10} = 2\).
2. Find the logarithm, \(\log_{10}(2) \approx 0.301\).
3. Add to \(pK_a\), \(pH = 4.76 + 0.301 = 5.061\), so the pH is about \(5.06\).

Tips and common mistakes

• Make sure \([A^-]\) and \([HA]\) use the same units, because the ratio only works correctly then.
• If \([A^-] = [HA]\), then \(\log_{10}(1) = 0\), so \(pH = pK_a\).

Frequently asked questions