Titration Calculations: Strong Acid - Strong Base

For a strong acid like hydrochloric acid, complete ionization occurs, so the hydronium ion concentration equals the acid's molarity. For example, 0.10 M HCl produces 0.10 M hydronium ions, giving a pH of 1.0. Simply use pH = -log[H+] to find the initial pH of the solution.

Before the equivalence point, excess acid remains in solution. Calculate the moles of hydronium ions remaining by subtracting moles of added hydroxide from initial moles of hydronium, then divide by total volume. For instance, adding 25 mL of 0.10 M NaOH to 50 mL of 0.10 M HCl leaves 0.0025 moles of hydronium in 75 mL, yielding pH 1.48.

At the equivalence point, all hydronium ions from the acid are neutralized by hydroxide ions from the base, producing only water and salt. Neither the cation nor anion undergoes acid-base ionization, so only water autoprotolysis generates hydronium ions. The pH equals 7.0, making the solution neutral.

Beyond the equivalence point, all acid is consumed and pH is determined by excess hydroxide ions. Calculate the moles of hydroxide remaining by subtracting initial moles of hydronium from total moles of hydroxide, then divide by total volume. Use pOH = -log[OH-] and pH = 14 - pOH to find the final pH.

Near the equivalence point, small additions of base cause large changes in hydronium ion concentration because the buffer capacity is minimal. Before the equivalence point, excess acid resists pH change, but as the equivalence point approaches, nearly all acid is consumed, making the solution sensitive to small base additions.

At the equivalence point, the moles of hydronium ions from the acid equal the moles of hydroxide ions from the base. For monoprotic strong acid and strong base, this means the moles of acid and base are equal. Using the equation moles = molarity × volume, you can determine the volume of titrant needed to reach equivalence.

The total volume in pH calculations includes both the initial acid solution and all added base titrant. For example, adding 25 mL of base to 50 mL of acid gives a total volume of 75 mL. This dilution effect is critical because it reduces the concentration of remaining hydronium or excess hydroxide ions, directly affecting the calculated pH.