Oxide and hydroxide interferences for Ba, lanthanides, Hf, and Ta

 

Barium and the lanthanides are relatively strong oxide formers. That is, ions like Ba+ and Ce+ react with oxygen in the plasma to form small proportions of BaO+ and CeO+ . The dominant masses of the oxide ions are 16 mass units higher than the isotope of the metal in the oxide. In the figure below, you can see that Ce has 4 naturally occurring isotopes (red bars) at masses 136, 138, 140, and 142. 140Ce is the most abundant isotope and is the one usually used for quantitative analysis. Pr has only 1 naturally occurring isotope, 141Pr (blue bars). These 5 M+ metal ions ions combine with 16O to form from 5 different mass oxide ions (MO+); for example: 136Ce+ becomes 136Ce16O+ at mass 152 (= 136 + 16).

 

 

Naturally there are also small contributions at other masses from 17O and 18O, which are 1 and 2 mass units higher than the 16O oxide. Hydroxides, such as 136Ce16OH+ at mass 153, can also be important. Neither of these is shown in Figure 1.

 

Figure 1 shows that CeO+ at mass 152 interferes with 152Sm analysis, and CeO+ at mass 158 interferes with 158Gd analysis. Oxides shown at masses 154, 156, and 157 are unimportant in this example because they are not used for quantitative analysis. There is no interference shown for 159Tb, although small quantities of 141Pr18O+, 142Ce17O+, and 142Ce16OH+ do interfere at this mass. Clearly for quantitative analysis of the metals, we must have a method of stripping away the interferences from the M+ signals.

 

 

The simplest way to determine the size of the interferences at the masses analyzed is to measure them directly by analysis of pure, single element solutions. Then, all you have to know is where to expect the interferences, see how big they are, and calculate how to subtract them from the total signal at each mass of interest. Table 1 shows just such a set of single solution analyses.

 

 

In Table 1, taking the ratio of the MO+ signals in the blue boxes to the M+ signals in the immediately overlying yellow boxes in Table 1 yields MO+/M+ ratios, shown in Table 2. The work needed to derive these MO+/M+ ratios is considerable, and so it is done rarely. The MO+/M+ ratios in Table 2 are referred to hereafter as the oxide calibration values.

 

 

Once we have the values in the table above, we can calculate the size of the oxide or hydroxide interference as follows:

 

 

Ioxide,cps = Intensity of the oxide interference (in cps) during actual quantitative analysis.

 

MO+cps,cal	= Oxide (or hydroxide) interference intensity (in cps) determined during oxide interference calibration (e.g., 142Ce16O+ at 158Gd+).
M+cps,cal	= M+ intensity (in cps) during oxide interference calibration at the mass used for quantitative analysis (e.g., 140Ce), determined during oxide interference calibration. Note that the mass used for quantitative analysis does not have to be the mass producing the interference.
M+cps,analyzed	= M+ intensity determined during actual quantitative analysis of unknown samples.

 

The [MO+cps,cal/M+cps,cal] ratios are therefore anempirical (and dimensionless) correction factors that can be used for unknown samples later on (Table 2). The absolute instrument sensitivity can vary while still preserving the proportion of oxides produced.

 

There is one problem with this method and that is that the MO+/M+ ratios are not constant; they vary with a wide variety of conditions including the purity of the argon carrier gas, the spray chamber temperature, the nebulizer gas flow, the nebulizer efficiency, and so on. Therefore, the correction factors themselves need adjustment each time lanthanides are analyzed. The graph below shows two calibration runs done to determine MO+/M+ as illustrated above. Although the two different runs had MO+/M+ ratios different by a factor of ~1.5, the changes in all of the interferences was approximately the same: ~1.5. As a result, all of the MO+/M+ ratios lie close to a straight line. Therefore, if we know the slope of this line on any given day during a run of unknowns, then we can correct the correction factors and so determine accurately the sizes of the interferences.

 

 

This is easily done by first analyzing a pure solution of a lanthanide, such as 146Nd+, and its oxide at 158Gd+. You may notice that 158 is not 16 mass units higher than 146.This is because the oxide at 158Gd+ is from 142Nd16O+ (mass 158), not from 146Nd16O+ (mass 162). It does not matter that our correction factor is derived from 146Nd+ rather than from the Nd isotope actually producing the oxide because the Nd isotope abundances are constant in nature (or nearly so).

 

What we need to do now is analyze a pure ~5 ppb Nd solution before running our unknowns, and measure the signal for 146Nd+ and 158Gd+ (= 142Nd16O+). This new 142Nd16O+/146Nd+ ratio will probably be different from the ratio determined during an earlier oxide calibration, as determined above.

 

 

C	= Correction factor, corrects all of today's MO+/M+ proportions to those measured during today's oxide calibration, according to the linear relationship shown in the figure above.
Numerator	= Today's determined NdO+/Nd+ ratio, determined at masses 158 and 146, respectively.
Denominator	= NdO+/Nd+ ratio, determined originally during the oxide calibration procedure. These values are also measured at masses 158 and 146, respectively.

 

Now we have the MO+/M+ proportions for all oxide and hydroxide interferences of Ba, the lanthanides, Hf, and Ta (determined some time ago during the oxide calibration), and we have the correction factor (determined today). During the analysis of unknowns, the signals on the analyzed element masses (yellow in Table 1) are used to calculate the oxide interference on all pertinent higher masses (blue in Table 1) by the equation:

 

 

These values can then be subtracted from the signals at the masses having the interferences to yield the actual signal produced by M+ ions, stripped of the oxide interferences:

 

 

Icorrected,cps	= Intensity of the M+ signal at a given mass, in cps, after removal of the interferences.
Ioriginal,cps	= Intensity of the total signal at a given mass, in cps, before removal of the interferences..

 

The ellipsis at the end of Equation 4 indicates additional interference correction like that in the curved brackets but for additional oxide species. For example, Nd and Sm produce oxides that interfere with 163Dy. Referring to Tables 1 and 2:

1. The interference from Nd on mass 163 is 23/56574 = 0.00041, based on the Table 1 oxide calibration.
2. The interference from Sm on mass 163 is 196/83404 = 0.00235, based on the Table 1 oxide calibration.

Using correction factor "C", which is determined each day using the pure Nd solution is constant for any given analysis procedure on a given day. The actual intensity of the mass 163 peak, due exclusively to 163Dy is:

 

163Dy = (original mass 163 intensity) - (C * 0.00041 * intensity of 142Nd) -(C * 0.00235 * intensity of 152Sm)  Equation 5.

 

These concepts are all well and good, but what do you need to really do before analyzing a bunch of samples?

 

 

Table 4 shows all of the oxide correction equations for Ba, the lanthanides, Hf, and Ta. The equations are more-or-less in the format that they appear in the Elan 6100 software, correction equations section. The coefficients vary over time as updates are made.

 

 

Although this oxide correction procedure is an approximation, it has worked quite well in practice. Note that there is a simpler way to enter the C correction in the Elan software, but in practice it does not work reliably.