Buffer Solutions: How to Prepare Them, Calculate pH, and Choose the Right Weak Acid/Base Pair
By ChemistryIQ Team · April 8, 2026
The Direct Answer: A Weak Acid Plus Its Conjugate Base Resists pH Changes
A buffer is a solution that resists significant changes in pH when small amounts of acid or base are added. The most common buffer recipe is a weak acid mixed with the salt of its conjugate base — for example, acetic acid (CH3COOH) mixed with sodium acetate (CH3COONa). When you add a small amount of strong acid like HCl, the conjugate base (acetate ion) neutralizes it, converting the added acid into more of the weak acid. When you add strong base like NaOH, the weak acid neutralizes it, converting to more conjugate base. In both cases, the pH barely changes because the buffer absorbs the insult.
The pH of any buffer is calculated with the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the molar concentration of the conjugate base, [HA] is the molar concentration of the weak acid, and pKa is the -log of the acid dissociation constant. When [A-] = [HA], the ratio is 1, log(1) = 0, and pH = pKa exactly. This is the most important relationship in buffer chemistry — the pKa of the weak acid tells you the optimal operating pH of the buffer.
Buffers work best when the target pH is within ±1 unit of the pKa. Outside that range, the buffer loses capacity because one component (either the acid or the conjugate base) becomes too scarce to neutralize additional acid or base. If you need a buffer at pH 7.4 (physiological), you pick a weak acid with pKa near 7.4 — phosphate buffer (H2PO4-/HPO4^2-, pKa 7.2) is the standard choice for biological work. For pH 4.7, acetate buffer (pKa 4.74) is perfect. For pH 9.2, ammonium/ammonia (pKa 9.25) is the match.
Snap a photo of any buffer problem and ChemistryIQ identifies the weak acid/base pair, calculates the pH using Henderson-Hasselbalch, and generates preparation instructions for hitting a target pH with given concentrations.
How Buffers Actually Work: The Chemistry
To understand buffers, focus on the equilibrium between a weak acid and its conjugate base: HA ⇌ H+ + A-. In a pure weak acid solution, very little dissociation occurs — that is why it is called weak. Adding the salt of the conjugate base (like sodium acetate) shifts the system so that both HA and A- are present in substantial concentrations. Now the solution can respond to either direction of disturbance.
Add a strong acid (H+) to the buffer. The added H+ reacts with the conjugate base: H+ + A- → HA. The strong acid is converted into more of the weak acid. The pH drops only slightly because the weak acid barely dissociates, so it does not contribute many additional H+ ions. The buffer absorbed the strong acid by sacrificing some of its A- pool.
Add a strong base (OH-) to the buffer. The added OH- reacts with the weak acid: OH- + HA → A- + H2O. The strong base is converted into water plus more conjugate base. The pH rises only slightly because the OH- was neutralized by the weak acid rather than adding directly to the free OH- pool. The buffer absorbed the strong base by sacrificing some of its HA pool.
The buffer capacity — how much acid or base it can absorb before the pH changes significantly — depends on two factors: (1) the total concentration of the buffer components (more concentrated buffer = more capacity), and (2) the ratio of HA to A- (closest to 1:1 = highest capacity). A buffer at pH = pKa has maximum capacity because both components are at equal concentration and it can absorb equally in both directions.
The buffer fails when added acid or base exceeds the pool of the component being consumed. Add enough HCl to react with ALL the A- and you no longer have a buffer — just a solution of weak acid + strong acid, which behaves like a regular weak acid solution. The pH can drop dramatically past this breakdown point. This is why buffers have a finite capacity and cannot absorb unlimited amounts of acid or base.
ChemistryIQ visualizes the equilibrium shift as buffer capacity is consumed and calculates exactly when the buffer will fail based on the amounts present.
Calculating Buffer pH With Henderson-Hasselbalch
The Henderson-Hasselbalch equation is the workhorse calculation for buffer chemistry: pH = pKa + log([A-]/[HA]).
Worked example 1: A buffer contains 0.20 M acetic acid (pKa = 4.74) and 0.30 M sodium acetate. Calculate the pH. pH = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92. The buffer is slightly above the pKa because more conjugate base (deprotonated form) shifts the ratio toward higher pH.
Worked example 2: A phosphate buffer contains 0.10 M KH2PO4 (weak acid, pKa = 7.20) and 0.15 M K2HPO4 (conjugate base). Calculate the pH. pH = 7.20 + log(0.15/0.10) = 7.20 + log(1.5) = 7.20 + 0.176 = 7.38. Very close to physiological pH (7.40) — which is why phosphate buffer is the standard for biochemistry labs.
Worked example 3 (back-calculation): You need a buffer at pH 5.00 using acetic acid and sodium acetate. What ratio of [A-] to [HA] is required? 5.00 = 4.74 + log([A-]/[HA]) 0.26 = log([A-]/[HA]) [A-]/[HA] = 10^0.26 = 1.82 You need a 1.82:1 ratio of acetate to acetic acid. If you use 0.10 M acetic acid, you need 0.182 M sodium acetate. Any pair of concentrations with this ratio produces pH 5.00 (within the limits of buffer capacity).
A common exam variation: instead of giving you M concentrations, the question gives you moles or grams of each component in a fixed volume. The ratio stays the same because both concentrations divide by the same volume. Example: 0.05 mol weak acid + 0.08 mol conjugate base in 250 mL gives the same pH as 0.10 M + 0.16 M because only the RATIO matters, not the absolute concentrations. The absolute concentrations only affect buffer CAPACITY, not pH.
The Henderson-Hasselbalch equation assumes the amounts you add are close to the equilibrium amounts — in other words, the weak acid dissociation and the conjugate base protonation are small enough to ignore. This assumption holds when both components are present in substantial amounts (10x or more than the ionization). For very dilute buffers, the assumption breaks and you need to solve the full ICE table.
ChemistryIQ solves buffer problems in both directions — forward (given concentrations, find pH) and backward (given target pH, find required ratio and amounts) — with step-by-step Henderson-Hasselbalch work.
Preparing a Buffer in the Lab: The Recipe Approach
For a wet-lab buffer preparation, you choose your weak acid based on the target pH, then calculate the ratio of HA to A- needed to hit that pH, then weigh out the appropriate amounts of each component and dissolve in water.
Step 1: Pick a weak acid with pKa within ±1 of your target pH. Target pH 4.7? Use acetic acid (pKa 4.74). Target pH 7.2? Use phosphate buffer (KH2PO4/K2HPO4, pKa 7.20). Target pH 9.0? Use ammonium chloride/ammonia (pKa 9.25) or Tris (pKa 8.07, usable up to about 9.0). Target pH 10.5? Use sodium bicarbonate/sodium carbonate (pKa 10.33).
Step 2: Use Henderson-Hasselbalch to find the required ratio. Target pH - pKa = log([A-]/[HA]). Invert the log to find the ratio directly.
Step 3: Decide on total buffer concentration. Typical biological buffers are 10-100 mM total. More concentrated buffers have more capacity but may interfere with other lab measurements. 25 mM is a common default for enzyme work.
Step 4: Calculate the actual amounts of each component. If you need 500 mL of 25 mM phosphate buffer at pH 7.4: total moles of buffer = 0.025 M × 0.500 L = 0.0125 mol. For pH 7.4 with pKa 7.20, the ratio [A-]/[HA] = 10^(7.4-7.2) = 10^0.2 = 1.585. So for every 1 mole of weak acid (H2PO4-), you need 1.585 moles of conjugate base (HPO4^2-). Solving: moles HA = 0.0125 / (1 + 1.585) = 0.00484 mol, moles A- = 0.0125 - 0.00484 = 0.00766 mol. Weigh out 0.658 g KH2PO4 (MW 136.09) and 1.335 g K2HPO4 (MW 174.18), dissolve in water, and top up to 500 mL.
Step 5: Verify the pH with a pH meter. Calculations give you a theoretical pH but real solutions may deviate slightly due to activity coefficients, ionic strength effects, and impurities. Adjust with small amounts of the weak acid or conjugate base until the meter reads the target pH.
Practical alternative (the shortcut method): instead of calculating exact amounts, make a concentrated stock of the weak acid and a stock of the conjugate base at the same molar concentration. Mix them together while monitoring pH with a meter until you hit the target. This is faster and more reliable than dry calculation because it accounts for real-world deviations from theoretical behavior.
Most biology labs keep premixed stock solutions of common buffers (PBS at pH 7.4, Tris at pH 8.0, citrate at pH 6.0) and dilute as needed. Making buffers from scratch every time is inefficient, so learn the standard recipes for your field.
ChemistryIQ generates buffer preparation protocols with exact mass calculations for target pH and volume — snap a photo of a problem and it walks through the calculations and tells you how many grams of each salt to weigh.
FAQs
Common questions about buffer solutions
The Ka expression (Ka = [H+][A-]/[HA]) is the fundamental equilibrium law. The Henderson-Hasselbalch equation is a rearranged version of it, obtained by taking the -log of both sides and using the definitions pH = -log[H+] and pKa = -log(Ka). Both describe the same chemistry. Henderson-Hasselbalch is more convenient for buffer calculations because it relates pH directly to the ratio of [A-] to [HA], which is what you care about in buffer problems. For dilute or unusual buffers, solving the full Ka expression with an ICE table may be more accurate.
Yes. Snap a photo of any buffer problem and ChemistryIQ identifies the weak acid/base pair, selects the correct pKa, applies Henderson-Hasselbalch in either direction (pH → ratio or ratio → pH), and generates preparation instructions including exact masses of each salt to weigh out for a target volume and concentration. It handles buffer capacity calculations, strong acid/base additions to existing buffers, and the common exam variations.