Buffer pH Calculator
Our mixtures & solutions calculator computes bufferp hcalculator accurately. Enter measurements for results with formulas and error analysis.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
pH = pKa + log([A-]/[HA])
The Henderson-Hasselbalch equation calculates buffer pH from the pKa of the weak acid and the logarithm of the ratio of conjugate base [A-] to weak acid [HA] concentrations.
Worked Examples
Example 1: Acetate Buffer
Problem:Calculate the pH of a buffer made from 0.1 M acetic acid and 0.15 M sodium acetate (pKa = 4.76).
Solution:pH = pKa + log([A-]/[HA])\npH = 4.76 + log(0.15/0.1)\npH = 4.76 + log(1.5)\npH = 4.76 + 0.176 = 4.936
Result:pH = 4.94
Example 2: Phosphate Buffer
Problem:Mix 50 mL of 0.2 M H2PO4- with 30 mL of 0.2 M HPO4^2- (pKa = 7.2). Find the pH.
Solution:Moles acid = 0.2 * 0.050 = 0.01 mol\nMoles base = 0.2 * 0.030 = 0.006 mol\npH = 7.2 + log(0.006/0.01)\npH = 7.2 + log(0.6) = 7.2 - 0.222 = 6.978
Result:pH = 6.98
Frequently Asked Questions
How do I choose the right buffer for my experiment?
Select a buffer whose pKa is within 1 pH unit of your target pH, because buffers work most effectively in this range. Common buffer systems include acetate (pKa 4.76) for pH 3.8-5.8, phosphate (pKa 7.2) for pH 6.2-8.2, and Tris (pKa 8.1) for pH 7.1-9.1. Consider whether the buffer components might interfere with your experiment, such as phosphate buffers precipitating calcium ions. Also check temperature sensitivity, as Tris buffer pH changes significantly with temperature while phosphate is more stable.
Why does a buffer resist pH changes?
A buffer resists pH changes because it contains both a weak acid (HA) and its conjugate base (A-) in significant amounts. When strong acid is added, the conjugate base neutralizes it: A- + H+ becomes HA. When strong base is added, the weak acid neutralizes it: HA + OH- becomes A- + H2O. This consumes the added acid or base without significantly changing the [A-]/[HA] ratio, keeping the pH relatively stable. The buffer fails when one component is depleted, which is why buffer capacity depends on concentration.
References
Reviewed by Manoj Kumar, Mathematics Educator ยท Editorial policy