Question

$\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ and $\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ }$ may be distinguished by treating their aqueous solutions with:
A) dil. Acetic acid
B) $\text{ MgO }$
C) Phenolphthalein
D) None

Answer 1

Sodium carbonate and sodium bicarbonate are basic in nature. Sodium carbonate when dissolved in water produces carbonic acid and sodium hydroxide. However, sodium bicarbonate is a very weak base compared to sodium carbonate.

Complete Solution :
Sodium carbonate is a disodium salt carbonic acid. When sodium carbonate $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ is dissolved in water it forms generates carbonate ions and hydroxide ions. When dissolved in water the carbonate ion forms carbonic acid and sodium hydroxide. The reaction of the sodium carbonate in an aqueous solution is as follows:
$\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ (aq) + }{{\text{H}}_{\text{2}}}\text{O }\to \text{ NaOH + }{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$
The amount of $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ content in a given sample is determined titration method.
- Indicator changes colour in well-defined $\text{ pH }$ range . Phenolphthalein changes its colour from colourless to pink from acidic to basic medium. Sodium carbonate $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ is titrated against the hydrochloric acid in presence of a phenolphthalein indicator. Sodium carbonate produces $\text{ NaOH }$ in solution. The phenolphthalein indicator turns pink.
- The aqueous solution of sodium carbonate and sodium bicarbonate are weakly basic in nature due to hydrolysis. But sodium bicarbonate is weaker than sodium carbonate ,Therefore it does not turn the phenolphthalein indicator to pink.
But sodium carbonate is relatively more basic than sodium bicarbonate and turns the solution to pink when phenolphthalein is added.
$\begin{aligned} & \text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3 }}}\text{+ Phenolphthalein indicator }\to \text{ Pink colour} \\\ & \text{ NaHC}{{\text{O}}_{\text{3 }}}\text{+ Phenolphthalein indicator }\to \text{ Colourless } \\\ \end{aligned}$
Therefore, the phenolphthalein indicator is used to distinguish sodium carbonate and sodium bicarbonate. So, the correct answer is “Option C”.

Note: Note that the sodium carbonate and sodium bicarbonate can be distinguished by heat. One important chemical test to distinguish between the $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ and $\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ }$ is $\text{ MgS}{{\text{O}}_{\text{4}}}\text{ }$ a test. The magnesium sulphate reacts with sodium carbonate to give magnesium carbonate.
- $\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + M}{{\text{g}}^{\text{2+}}}\to \text{ MgC}{{\text{O}}_{\text{3}}}\text{ }\downarrow \text{(white) + N}{{\text{a}}^{\text{+}}}$
- Sodium bicarbonate when heated in presence of $\text{ MgS}{{\text{O}}_{\text{4}}}\text{ }$ forms white colour salt magnesium carbonate along with the evolution of carbon dioxide. The gas is detected by passing through the lime water.
-$\text{ NaHC}{{\text{O}}_{\text{3}}}\text{ + M}{{\text{g}}^{\text{2+}}}\to \text{ MgC}{{\text{O}}_{\text{3}}}\text{ }\downarrow \text{(white) + }{{\text{H}}_{\text{2}}}\text{O + C}{{\text{O}}_{\text{2}}}\uparrow \text{ }$