Qualitative Analysis - Result Question 13

####13. A sodium salt of an unknown anion when treated with MgCl2 gives white precipitate only on boiling. The anion is

(a) SO42−

(b) HCO3−

(c) CO32−

(d) NO3−

(2004,1M)

Show Answer

Answer:

Correct Answer: 13. (b)

Solution:

  1. A sodium salt of an unknown anion when treated with MgCl gives white precipitate (MgCO3) only on boiling. Hence, the action must be HCO3−ion.

MgCl2+2NaHCO3→ΔMgCO3+2NaCl+H2O+CO2

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