= Primary standard inventory
{title2=$n_s=mp/M$}
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,
$$
n_s=\frac{mp}{M},\qquad c_B=\frac{rmp}{MV}.
$$
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> $M=204.22\,\mathrm{g\,mol^{-1}}$. The NIST Chemistry WebBook lists formula $\mathrm{C_8H_5KO_4}$ and molecular weight $204.2212$, consistent with the rounded <molar mass> used here.<NIST Chemistry WebBook, SRD 69/potassium hydrogen phthalate>[NIST Chemistry WebBook, SRD 69, potassium hydrogen phthalate, CAS 877-24-7; Formula and Molecular weight fields]{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> is
$$
\mathrm{HC_8H_4O_4^-}+\mathrm{OH^-}\longrightarrow\mathrm{C_8H_4O_4^{2-}}+\mathrm{H_2O}.
$$
Thus <sodium hydroxide> has one-to-one <stoichiometry> with <potassium hydrogen phthalate> in this <neutralization>, giving
$$
n_s=\frac{0.51055\,\mathrm g}{204.22\,\mathrm{g\,mol^{-1}}}=0.002500\,\mathrm{mol}.
$$
Choose a delivered endpoint <titre> of $24.85\,\mathrm{mL}$ and matched <titration blank> of $0.05\,\mathrm{mL}$. Then
$$
V=(24.85-0.05)\,\mathrm{mL}=0.02480\,\mathrm L,\qquad
c_B=\frac{0.002500\,\mathrm{mol}}{0.02480\,\mathrm L}
=0.1008064516\ldots\,\mathrm{mol\,L^{-1}}.
$$
The extra displayed digits show the arithmetic of the stipulated model; they do not establish experimental accuracy. A compact rounded result is $0.1008\,\mathrm{mol\,L^{-1}}$.
\Image[/primary-standard-mass-amount-titre-concentration.png]
{title=A <primary standard> connects known <mass> to <amount of substance>, corrected <titre>, and <titrant> <concentration>}
{description=Original calculated inventory diagram for selected $0.51055\,\mathrm g$ <potassium hydrogen phthalate>, stipulated $p=1$, rounded <molar mass> $204.22\,\mathrm{g\,mol^{-1}}$, and one-to-one <sodium hydroxide> <neutralization>. The $24.85\,\mathrm{mL}$ endpoint and $0.05\,\mathrm{mL}$ matched <titration blank> are chosen examples, not observations.}
When the entire weighed portion is titrated, added <water> changes <concentration> and the <pH> path but preserves the initial acid inventory. The dissolution <volume (thermodynamics)> 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 $n_sV_a/V_f$ for homogeneous stock, with delivered <aliquot> <volume (thermodynamics)> $V_a$ and final stock <volume (thermodynamics)> $V_f$. The same bookkeeping explains why complete transfer matters even when the amount of dissolution <water> is approximate.
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