The titration with base gives us two measured quantities from which we can determine the composition of a mixture of the two components, SA and ASA.
The weight of the sample titrated, W, must be related to the number of moles of SA and ASA, nSA and nASA respectively, by:
W = 138.12 nSA + 180.15 nASA
[ If there were a third component (whether it is an acid or not, e.g., water), the above relationship would not be correct and the rest of the analysis below would not be correct.]
Since each acid is monoprotic, the number of moles of titratable hydrogen ion in the mixture must be equal to the number of moles of NaOH used to titrate the sample, N, which must be:
N = nSA + nASA
Thus we have two equations in the unknowns nSA and nASA which can be solved to give:
(180.15 N– W) (W – 138.12 N)
nSA = ---------------------------- and nASA = ----------------------------
(180.15 – 138.12) (180.15 – 138.12)
[ You should be able to derive the above two formulas.]
Dividing by the total number of moles of acid in the sample, N, and multiplying by 100 to convert to percentages gives us:
100 (180.15 – W/N) 100 (W/N– 138.12)
%SA = ---------------------------- and %ASA = --------------------------
(180.15 – 138.12) (180.15 – 138.12)
But, W/N is exactly what we can call the effective molar mass, EMM, hence
100 (180.15 – EMM) 100 (EMM – 138.12)
%SA = ---------------------------- and %ASA = ----------------------------
(180.15 – 138.12) (180.15 – 138.12)
These are precisely the formulas shown in SUSB-012.
Last Update: 27-Apr-2006