A solution of a mixture is made alkaline with NH4OH in the presence of NH4Cl and treated with (NH4)2CO3 (commercial ammonium carbonate contains bicarbonate/carbamate). Which statements correctly explain the procedure?

Excess NH3 converts HCO3- to CO3^2- and raises the carbonate available for precipitation.

NH4Cl keeps [CO3^2-] low enough that MgCO3 does not precipitate.

Excess NH3 converts HCO3- to CO3^2- and raises the carbonate available for precipitation.

NH4Cl keeps [CO3^2-] low enough that MgCO3 does not precipitate.

NH4Cl makes the medium acidic enough to dissolve BaCO3 as it forms.

The group V precipitates are coloured (yellow to orange) carbonates.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3qc148

A solution of a mixture is made alkaline with $NH X_{4} OH$ in the presence of $NH X_{4} Cl$ and treated with $(NH X_{4}) X_{2} CO X_{3}$ (commercial ammonium carbonate contains bicarbonate/carbamate). Which statements correctly explain the procedure?

AExcess $NH X_{3}$ converts $HCO X_{3} X^{-}$ to $CO X_{3} X^{2 -}$ and raises the carbonate available for precipitation.
B $NH X_{4} Cl$ keeps $[CO X_{3} X^{2 -}]$ low enough that $MgCO X_{3}$ does not precipitate.
C $NH X_{4} Cl$ makes the medium acidic enough to dissolve $BaCO X_{3}$ as it forms.
DThe group V precipitates are coloured (yellow to orange) carbonates.

HCO X_{3} X^{-} + NH X_{3} CO X_{3} X^{2 -} + NH X_{4} X^{+}

is driven right by excess ammonia, so more carbonate is available (A true). The ammonium ion limits

[CO X_{3} X^{2 -}]

so that only the less soluble carbonates precipitate and

Mg X^{2 +}

stays in solution (B true). The medium remains alkaline, and the carbonates of Ba, Sr, Ca are white.