Calculate the pH of mixtures containing strong acids (HCl, H₂SO₄, HNO₃) and strong bases (NaOH, KOH, Ba(OH)₂). Enter concentrations and volumes for each component, and see the resulting pH after neutralization.
Add strong acids, bases, and/or salts with their concentrations and volumes to calculate the pH after mixing.
If you're working through a simple pH of mixtures problem and your answer doesn't make sense, the most likely error is arithmetic with moles. The entire calculation hinges on one comparison: total moles of H⁺ from all acid sources versus total moles of OH⁻ from all base sources. Whichever is larger wins, and the leftover determines the final pH.
Start by computing moles for each component separately: moles = molarity × volume (in liters). If you have 50 mL of 0.10 M HCl, that's 0.050 L × 0.10 M = 0.0050 mol H⁺. If you also have 30 mL of 0.10 M NaOH, that's 0.030 L × 0.10 M = 0.0030 mol OH⁻. Sum all the acid moles on one side and all the base moles on the other.
Then subtract: net H⁺ = total H⁺ − total OH⁻. If net H⁺ is positive, the solution is acidic. If negative, there's excess OH⁻ and the solution is basic. If exactly zero, it's neutral. This subtraction is the neutralization step—H⁺ and OH⁻ react 1:1 to form water, so every mole of one cancels a mole of the other. Everything downstream depends on getting these moles right.
Once you know which species is in excess, the pH calculation splits into two paths. If H⁺ is left over: [H⁺] = excess moles H⁺ ÷ total volume, then pH = −log[H⁺]. If OH⁻ is left over: [OH⁻] = excess moles OH⁻ ÷ total volume, then pOH = −log[OH⁻] and pH = 14 − pOH. Using the wrong path is a guaranteed wrong answer.
The most common mistake is computing excess correctly but then dividing by the wrong volume. Total volume means the sum of every solution you mixed—not the volume of just one component. If you mixed 50 mL of acid with 30 mL of base, total volume is 80 mL = 0.080 L. Students who use 0.050 L or 0.030 L get concentrations that are too high or too low, throwing off the pH.
Watch the units throughout. Molarity is mol/L, so volume must be in liters. If you leave volume in mL and multiply by molarity, your moles are off by a factor of 1000. This cascading error is the single biggest source of wrong answers on mixing problems. Convert mL to L first, every time, before any multiplication.
Mixing two solutions together dilutes both of them, even when one is pure water. If you take 50 mL of 0.10 M HCl (with no base at all) and add 50 mL of water, you now have 0.0050 mol H⁺ in 100 mL. The concentration drops from 0.10 M to 0.050 M, and pH rises from 1.00 to 1.30. The acid doesn't get neutralized—it just spreads out.
Dilution shifts pH toward 7 from either direction. Acidic solutions become less acidic (pH rises) and basic solutions become less basic (pH drops). But dilution can never cross pH 7 or overshoot neutral. If you start at pH 2 and keep diluting, pH approaches 7 asymptotically—it never reaches 8. If your calculation gives an acidic solution turning basic just from dilution, something went wrong.
At extreme dilutions (below about 10⁻⁶ M), the autoionization of water (Kw = 10⁻¹⁴) starts mattering. The simple pH = −log[H⁺] formula breaks down because water's own H⁺ contribution is no longer negligible. For typical exam problems with concentrations above 10⁻⁴ M, you don't need to worry about this.
When moles of H⁺ exactly equal moles of OH⁻, complete neutralization occurs: every H⁺ pairs with an OH⁻ to form water. No excess of either species remains. For strong acid + strong base mixtures, the resulting solution is just salt and water, and pH = 7.00 at 25 °C.
This is the only scenario where the solution is perfectly neutral. Even a tiny imbalance—say 0.0001 mol excess H⁺ in 100 mL—gives [H⁺] = 0.001 M and pH = 3.00, which is noticeably acidic. Complete neutralization is a knife's edge. On an exam, if the moles come out exactly equal, state pH = 7 and explain why: Kw governs [H⁺] when no excess strong electrolyte is present.
Important: this pH = 7 result applies only to strong acid + strong base combinations. If the acid were weak (like acetic acid), equivalence would produce a solution of the conjugate base, and pH would be above 7. But this tool handles strong electrolytes only, where complete dissociation means complete neutralization gives a neutral solution.
Does mixing order matter? No. Whether you pour acid into base or base into acid, the moles of H⁺ and OH⁻ are the same, so the equilibrium result is identical. (In practice, adding acid to base is safer because the base can absorb the heat, but that's a safety consideration, not a calculation one.)
What about polyprotic acids? H₂SO₄ contributes 2 mol H⁺ per mole (both dissociations are effectively complete for strong-electrolyte purposes). So 0.01 mol H₂SO₄ provides 0.02 mol H⁺. Ba(OH)₂ gives 2 mol OH⁻ per mole. Always multiply moles of substance by the number of H⁺ or OH⁻ ions it releases before comparing acid to base totals.
Can I mix more than two solutions? Yes. Just sum all H⁺ contributions from every acid and all OH⁻ contributions from every base. The subtraction gives net excess, and you divide by the total combined volume. The same logic works for 2, 3, or 20 components.
Why doesn't this work for weak acids? Weak acids don't fully dissociate. If you mix acetic acid with NaOH, the neutralization produces acetate ion, which itself is a base. You'd need the Henderson–Hasselbalch equation or a Ka equilibrium calculation, not the simple excess-moles approach.
• Core reaction: H⁺ + OH⁻ → H₂O. One-to-one stoichiometry, no coefficients to adjust. The reaction goes to completion for strong electrolytes.
• Moles = M × V(L): Convert volume to liters first. This is the only formula you need for the moles step. Never mix units—0.05 M × 25 mL ≠ 1.25 mol.
• pH from excess H⁺: [H⁺] = excess mol / V_total(L). pH = −log[H⁺]. Use this when acid is in excess.
• pH from excess OH⁻: [OH⁻] = excess mol / V_total(L). pOH = −log[OH⁻]. pH = 14 − pOH. Use this when base is in excess.
• Neutral case: mol H⁺ = mol OH⁻. [H⁺] = 10⁻⁷ M from Kw. pH = 7.00 at 25 °C.
Problem: Mix 40.0 mL of 0.15 M HCl with 25.0 mL of 0.20 M NaOH. Find the pH.
Step 1: Calculate moles
mol H⁺ = 0.15 × 0.0400 = 0.00600 mol
mol OH⁻ = 0.20 × 0.0250 = 0.00500 mol
Step 2: Neutralize
Excess H⁺ = 0.00600 − 0.00500 = 0.00100 mol
Step 3: Total volume
V_total = 40.0 + 25.0 = 65.0 mL = 0.0650 L
Step 4: Concentration and pH
[H⁺] = 0.00100 / 0.0650 = 0.01538 M
pH = −log(0.01538) = 1.81
Sanity check: acid was in excess, so pH should be below 7—it is. The excess is small (0.001 mol in 65 mL), so pH isn't extremely low—1.81 is reasonable. If you'd accidentally used only 40 mL as total volume, you'd get [H⁺] = 0.025 M and pH = 1.60—wrong by 0.21 pH units, enough to lose marks.