Free tool
Find the pH of a buffer from the pKa (or Ka) and the amounts of conjugate base and weak acid, or work backward to the base-to-acid ratio a target pH needs. It flags any ratio outside the 1:10 to 10:1 buffer range.
Built for General Chemistry IIScientific notation works: 1.8e-5. For a weak-base buffer such as NH3/NH4+, enter the pKa of the conjugate acid (14.00 - pKb at 25 °C).
Enter the acid constant and amounts.
Buffer questions show up on every Gen Chem II exam. Encodr turns pKa, the buffer range and the added-acid steps into spaced-repetition practice so they hold until the final.
Get started freepH = pKa + log([A-] / [HA]), where [A-] is the conjugate base, [HA] is the weak acid and
pKa = -log Ka. Because both species sit in the same volume, you can use moles instead of molarity:
only the ratio matters. Solving for the ratio gives [A-] / [HA] = 10^(pH - pKa).
When [A-] = [HA], log(1) = 0 and pH = pKa. That 1:1 point is where the buffer resists change best. A buffer works well while the ratio stays between about 1:10 and 10:1, which is a pH within 1 unit of the pKa.
Worked example: 0.20 M acetic acid (Ka = 1.8 x 10^-5, pKa 4.74) with 0.30 M sodium acetate. pH = 4.74 + log(0.30 / 0.20) = 4.74 + 0.18 = 4.92. Add 0.020 mol NaOH to 1.00 L of it: the base converts 0.020 mol HA to A-, giving 0.32 mol A- and 0.18 mol HA, so pH = 4.74 + log(0.32 / 0.18) = 4.99. More examples are in the Henderson-Hasselbalch equation and buffers, and the whole course is in the free General Chemistry II deck.
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