A) \[M{{g}^{2+}}\] ions will also be precipitated.
B) Concentration of\[C{{O}_{3}}^{2-}\] 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 \[{{(N{{H}_{4}})}_{2}}C{{O}_{3}}.\]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 \[-N{{H}_{2}}\] 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.
You need to login to perform this action.
You will be redirected in
3 sec
