Mix several acids, bases and salts —strong or weak, monoprotic or polyprotic— with their molarity and volume, and get the pH, the pOH and the concentration and fraction of each species at equilibrium. Choose which one is added from the burette and you get the titration curve with the equivalence points. The exact charge balance is solved, without approximations.
Atomic masses, ionization energies, melting points and names from Wikidata, public-domain data (CC0 1.0), checked one by one by Simuveliq. Atomic masses are the IUPAC standard ones.
pH measures the acidity of an aqueous solution: pH = −log[H3O+]. Computing it is easy for strong acids and bases, but weak acids, buffers, salts or mixtures require taking the equilibria into account. This guide reviews the key ideas, works the typical cases and explains how the calculator solves them without approximations.
For acetic acid, pKa = 4.756 (Ka = 1.75 × 10⁻⁵) is used, the value the tool provides.
1. Strong acid. 0.010 M HCl: [H3O+] = 0.010 M and pH = 2.00.
2. Weak acid. 0.10 M acetic acid. With the usual approximation, x = [H+] ≈ √(Ka·C) = √(1.75 × 10⁻⁶) ≈ 1.32 × 10⁻³ M, so pH ≈ 2.88. The approximation holds because x is only 1.3 % of C. The exact calculation gives pH = 2.881.
3. Buffer. 50 mL of 0.10 M acetic acid are mixed with 25 mL of 0.10 M NaOH. The base neutralizes half of the acid, leaving equal amounts of acid and acetate. By the Henderson–Hasselbalch equation,
pH = pKa + log([A−]/[HA]) = 4.756 + log 1 = 4.76
4. Salt of a weak acid. 0.050 M sodium acetate: Kb = Kw/Ka ≈ 5.7 × 10⁻¹⁰; [OH−] ≈ √(5.7 × 10⁻¹⁰ · 0.050) ≈ 5.3 × 10⁻⁶ M; pOH ≈ 5.27 and pH ≈ 8.73. A salt solution can be basic even though it contains no "strong" base.
5. A classic trap. 1.0 × 10⁻⁸ M HCl does not have pH 8: an acid cannot make water basic. Here the H+ from water matters more than the acid's, and the exact calculation gives pH ≈ 6.98.
The calculator gives these same values.
Instead of picking an approximate formula for each case, the tool writes the charge balance (total positive charge equals total negative charge) together with the mass balances of each acid–base system. The fractions α of each species depend only on pH, so one equation in one unknown remains, which is solved by bisection. That way it works the same for polyprotic acids, mixtures of several acids and bases, buffers or salts, and in example 5 it gives the right answer with no tricks. It assumes an ideal solution (no activity coefficients), 25 °C and additive volumes.
D. C. Harris, Quantitative Chemical Analysis (W. H. Freeman); R. Chang and K. A. Goldsby, Chemistry (McGraw-Hill). Simuveliq is not affiliated with these authors or publishers.
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