 Primary standard inventory

Let $m$ be the transferred sample mass, $p$ the known mass fraction of the intended chemical compound, and $M$ its molar mass. In a model with complete dissolution and chemically inactive impurities,The purity fraction is dimensionless: $99.95\%$ would mean $p=0.9995$, not $99.95$. These equations assume that the impurity contributes no titrant demand; an impurity that also reacts needs its own inventory equation. The value of $p$ must refer to the same preparation state as $m$, for example the specified dry material. A mass fraction assigned before unaccounted uptake of water cannot simply be used after it.

For an original worked example, stipulate pure potassium hydrogen phthalate, abbreviated KHP, so $p=1$, a selected mass $m=0.51055\mathrm{g}$, and rounded molar mass $M=204.22\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The NIST Chemistry WebBook lists formula ${\mathrm{C}}_8{\mathrm{H}}_5\mathrm{K}{\mathrm{O}}_4$ and molecular weight $204.2212$, consistent with the rounded molar mass used here.^[ref] The selected mass, purity assumption and volumes in this example are chosen inputs, not measured results or a claim about a real batch's purity.

The acid-bearing anion of potassium hydrogen phthalate loses one proton to hydroxide. The net chemical equation isThus sodium hydroxide has one-to-one stoichiometry with potassium hydrogen phthalate in this neutralization, givingChoose a delivered endpoint titre of $24.85\mathrm{mL}$ and matched titration blank of $0.05\mathrm{mL}$. ThenThe extra displayed digits show the arithmetic of the stipulated model; they do not establish experimental accuracy. A compact rounded result is $0.1008\mathrm{mol}{\mathrm{L}}^{-1}$.

When the entire weighed portion is titrated, added water changes concentration and the pH path but preserves the initial acid inventory. The dissolution volume therefore does not appear in $n_s=mp/M$. If instead a volumetric flask prepares a stock and a volumetric pipette selects an aliquot, the selected inventory is $n_s{V}_a/V_f$ for homogeneous stock, with delivered aliquot volume $V_a$ and final stock volume $V_f$. The same bookkeeping explains why complete transfer matters even when the amount of dissolution water is approximate.