Question

Chlorate $\left({{\mathrm{CIO}}^{-}}_3\right)$, chlorite $\left({{\mathrm{CIO}}^{-}}_2\right)$, bromate $\left({{\mathrm{BrO}}^{-}}_3\right)$, and iodate $\left({{\mathrm{IO}}^{-}}_3\right)$can be measured in drinking water at the 1-ppb level with 1% precision by selected reaction monitoring. Chlorate and chlorite arise from ${\mathrm{CIO}}_2$used as a disinfectant. Bromate and iodate can be formed from ${\mathrm{Br}}^{-}$or ${\mathrm{I}}^{-}$when water is disinfected with ozone $\left({\mathrm{O}}_3\right)$. For the highly selective measurement of chlorate, the negative ion selected by Q1 in Figure 22-33 is m/z 83 and the negative ion selected by Q3 is m/z 67. Explain how this measurement works and how it distinguishes ${\mathrm{CIO}}_3^{-}$from ${\mathrm{CIO}}_2^{-}$, ${\mathrm{BrO}}_3^{-}$, and${\mathrm{IO}}_3^{-}$

Short answer

The answer is not given in the drive.

Step-by-step solution

Test CIO3-:

Test whether${\mathbf{CIO}}_{\mathbf{3}}^{\mathbf{-}}$can be distinguished from all three other compound in drinking water, the m/z of all of the compounds should be computed.

${\mathrm{CIO}}_2^{-}$ has m/z of 67 which pass through third quadrupole, it wouldn’t have pass through second quadrupole.

further explanation:

When ${\mathrm{CIO}}_3^{-}$ passes through the second quadrupole, it undergoes collisionally activated dissociation, breaking into fragments. One of the possible fragments to be formed is${\mathrm{CIO}}_2^{-}$ fragment. The only fragment that would be able to leave the third quadrupole, reaching the detector.