Analytical Chem istry - DePauw University

168 Analytical Chemistry 2.0have a sample and a standard with identical concentrations of analyte andidentical signals. If both experience the same proportional loss of solventthen their respective concentrations of analyte and signals continue to beidentical. In effect, we can ignore evaporation if the samples and standardsexperience an equivalent loss of solvent. If an identical standard and samplelose different amounts of solvent, however, then their respective concentrationsand signals will no longer be equal. In this case a simple externalstandardization or standard addition is not possible.We can still complete a standardization if we reference the analyte’ssignal to a signal from another species that we add to all samples and standards.The species, which we call an internal standard, must be differentthan the analyte.Because the analyte and the internal standard in any sample or standardreceive the same treatment, the ratio of their signals is unaffected by anylack of reproducibility in the procedure. If a solution contains an analyte ofconcentration C A , and an internal standard of concentration, C IS , then thesignals due to the analyte, S A , and the internal standard, S IS , areS= k CA A AS= k CIS IS ISwhere k A and k IS are the sensitivities for the analyte and internal standard.Taking the ratio of the two signals gives the fundamental equation for aninternal standardization.SSAISkCKC A AA= = × 5.12kC CISBecause K is a ratio of the analyte’s sensitivity and the internal standard’ssensitivity, it is not necessary to independently determine values for eitherk A or k IS .Si n g l e In t e r n a l St a n d a r dIn a single-point internal standardization, we prepare a single standard containingthe analyte and the internal standard, and use it to determine thevalue of K in equation 5.12.C SK = ⎛ ⎞× ⎛ ⎞ISA5.13⎝⎜C ⎠⎟⎝⎜S ⎠⎟A std IS stdHaving standardized the method, the analyte’s concentration is given byC SIS AC = × ⎛ ⎞AK ⎝⎜S ⎠⎟ISISsampIS