Correct Answer - B
Unlike `NaHCO_(3)` which is sparingly soluble in water, `KHCO_(3)` is fairly soluble in water. Thus, when `CO_(2)` is passed through ammoniated brine, `NaHCO_(3)` gets precipitated, while `KHCO_(3)` does not get precipitated when `CO_(2)` is passed through an ammoniacal solution of KCl. However, it can be made by passing `CO_(2)` into a solution of potassium hydroxide. Evaporation and subsequent ingnition gives the corbonate:
`overset(+)(K)overset(-)(O)H(aq.)+CO_(2)(g)toK^(+)HCO_(3)^(-)(aq.)`
`2K^(+)HCO_(3)^(-)(aq.)to(K^(+))_(2)CO_(3)^(2-)(s)+H_(2)O(l)+CO_(2)(g)`.