Two solutions are labeled A and B. Solution A contains \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and solution \(\mathrm{B}\) contains \(\mathrm{NaHCO}_{3}\). Describe how you would distinguish between the two solutions if you were provided with a \(\mathrm{MgCl}_{2}\) solution. (Hint: You need to know the solubilities of \(\mathrm{MgCO}_{3}\) and \(\mathrm{MgHCO}_{3}\).)

Short Answer

Expert verified
The amount of precipitate in the solutions after the addition of the \(MgCl_2\) solution distinguishes the two solutions. Solution A that contains \(Na_2CO_3\) will have more precipitate due to the formation and poor solubility of \(MgCO_3\). In contrast, solution B that contains \(NaHCO_3\) will have less precipitate because of the relatively higher solubility of \(Mg(HCO_3)_2\) which is formed.

Step by step solution

01

Reaction of solution A and solution B with \(MgCl_2\)

First, separate the two solutions (A and B) into two distinct containers. Then add \(MgCl_2\) to both containers. This will create reactions which will produce new compounds. In solution A, the reaction will be: \(Na_2CO_3 + MgCl_2 → MgCO_3 + 2NaCl\). In solution B, the reaction will be: \(2NaHCO_3 + MgCl_2 → Mg(HCO_3)_2 + 2NaCl\)
02

Comparing the solutions

After the reactions, both solutions will have produced different compounds: \(MgCO_3\) and \(Mg(HCO_3)_2\), respectively. Both \(MgCO_{3}\) (Magnesium carbonate) and \(MgHCO_{3}\) (Magnesium bicarbonate) are insoluble in water. However, \(MgCO_{3}\) is less soluble than \(MgHCO_{3}\). This means that in solution A, there will be more sediments or precipitates compared to solution B due to the insolubility of \(MgCO_{3}\) in water.
03

Confirm the differences

Finally, confirm the results by observing the amount of precipitate (the solid that forms and settles out of a liquid mixture) in both solutions. The solution with more precipitate is solution A (containing \(Na_2CO_3\)), while the solution with less precipitate is solution B (containing \(NaHCO_3\)).

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