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  • How to Accurately Calculate the pH of a Two‑Chemical Mixture

    By Blake Flournoy | Updated Mar 24, 2022

    Image credit: StanislavSalamanov/iStock/GettyImages

    pH is the logarithmic measure of hydrogen ion concentration in a solution, ranging from 0 to 14. For single‑component solutions, calculating pH is straightforward. When two acids or two bases are combined, the calculation becomes slightly more involved, yet remains manageable with the right approach.

    TL;DR

    Mixing acids or bases produces a neutralization reaction that generates salt (and sometimes water). To find the pH of a two‑chemical mixture:

    • Determine the volume and molarity of each component.
    • Compute the total moles of H⁺ (for acids) or OH⁻ (for bases) introduced into the mixture.
    • Divide the total H⁺ concentration by the total volume of the mixed solution to get [H⁺].
    • Apply pH = –log₁₀[H⁺].

    Always wear appropriate PPE, especially when handling strong acids or bases.

    Understanding pH

    pH quantifies the concentration of hydrogen ions in a solution. While the term is most commonly associated with liquids, it is equally relevant for soils, gels, and other media. Since pH is a logarithmic scale, small changes in [H⁺] can produce significant shifts in acidity or alkalinity.

    Component Calculations

    Before combining two acids or two bases, gather the following data for each component:

    • Volume (in liters or milliliters)
    • Molarity (mol L⁻¹)

    Using this information, calculate the number of moles of hydrogen ions (for acids) or hydroxide ions (for bases) contributed by each solution. The sum of these moles represents the total ionic content of the mixture.

    Mixed Solutions

    Once the total moles of H⁺ are known, the concentration of hydrogen ions in the final mixture is:

     [H⁺] = (total moles H⁺) / (total volume in liters)

    For example, if two acids provide 0.025 mol and 0.015 mol of H⁺, respectively, and the combined volume is 200 mL (0.200 L), then:

     [H⁺] = (0.025 + 0.015) mol / 0.200 L = 0.0002 mol L⁻¹

    Finally, the pH is calculated as:

     pH = –log₁₀(0.0002) ≈ 3.70

    Repeat the analogous steps for base–base mixtures, substituting [OH⁻] and using the relation pH = 14 – pOH.

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