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Chapter 17: Problem 96

What reagents would you employ to separate these pairs of ions in solution: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+},\) (b) \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+},\) (c) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) ?

Short Answer

Expert verified
The reagents that can be used to separate the ions are: \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+}\) by H2SO4, \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+}\) by K2CrO4, \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) by KI.

Step by step solution

01

Na+ and Ba2+

Considering the pair of ions \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+},\) sulfuric acid (H2SO4) can be used. Sulfate (SO4^{2-}) forms a precipitate with \(\mathrm{Ba}^{2+}\) (BaSO4), but not with sodium ion \(\mathrm{Na}^{+}\). Therefore, adding sulfuric acid will result in the precipitation of BaSO4, leaving Na+ in solution.
02

K+ and Pb2+

For the pair of ions \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+},\) potassium chromate (K2CrO4) can be used. The chromate anion (CrO4^{2-}) forms a yellow precipitate with lead ion \(\mathrm{Pb}^{2+}\) (PbCrO4), but not with \(\mathrm{K}^{+}\). In this case, adding potassium chromate will precipitate PbCrO4, leaving K+ in solution.
03

Zn2+ and Hg2+

The last pair consists of \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\). Potassium iodide (KI) can be used. The iodide anion (I-) forms a precipitate with \(\mathrm{Hg}^{2+}\) (HgI2), but not with \(\mathrm{Zn}^{2+}\). Hence, the addition of potassium iodide will precipitate HgI2, leaving \(\mathrm{Zn}^{2+}\) in solution.

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Most popular questions from this chapter

The \(\mathrm{p} K_{\mathrm{a}}\) of the indicator methyl orange is \(3.46 .\) Over what pH range does this indicator change from \(90 \%\) HIn to \(90 \% \mathrm{In}^{-} ?\)

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