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In fifth group, ${({NH}_{4})}_{2} {CO}_{3}$ is added to precipitate out the carbonates. We do not add ${Na}_{2} {CO}_{3}$ because:

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${MgCO}_{3}$ will be precipitated out in fifth group

${Na}_{2} {CO}_{3}$ increases the solubility of fifth group carbonates

none of the above

${CaCO}_{3}$ is soluble in ${Na}_{2} {CO}_{3}$

answer is C.

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Detailed Solution

As ${Na}_{2} {CO}_{3}$ is a strong electrolyte it will cause the precipitation of ${MgCO}_{3} \cdot$ So, ${({NH}_{4})}_{3} {CO}_{3}$ is used as it is a weaker electrolyte than ${Na}_{2} {CO}_{3}$ .

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