A) \[M{{g}^{2+}}\] ions will also be precipitated.
B) Concentration of \[CO_{3}^{2-}\] ions is very low.
C) Sodium ions will react with acid radicals.
D) \[N{{a}^{+}}\] ions will interfere with the detection of \[C{{a}^{2+}},B{{a}^{2+}},S{{r}^{2+}}\] ions.
Correct Answer: A
Solution :
[a] If \[N{{a}_{2}}C{{O}_{3}}\] is used in place of\[{{(N{{H}_{4}})}_{2}}C{{\text{O}}_{3}}\]. It will precipitate group V radicals as well as magnesium radicals. The reason for this is the high ionization of \[N{{a}_{2}}C{{O}_{3}}\] in water into \[N{{a}^{+}}\] and\[CO_{3}^{2-}\]. Now the higher concentration of \[CO_{3}^{2-}\] is available which exceeds the solubility product of group V radicals as well as that of magnesium radicals.
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