Shall we prepare and see?

Chemistry Level 2

Among $\ce{Na_2CO_3}$ and $\ce{K_2CO_3}$ , which can be easily prepared?

\ce{Na_2CO_3}

\ce{K_2CO_3}

1 solution

Skanda Prasad
Jan 28, 2017

Solubility of I A group of elements are in the following order:

$Li_2CO_3<Na_2CO_3<K_2CO_3<Rb_2CO_3<Cs_2CO_3$

We can observe that $Na_2CO_3$ is less soluble in water than $K_2CO_3$ .

$\implies$ It is easier to prepare $Na_2CO_3$

$K_2CO_3$ is formed with the reaction of $KOH$ and $H_2CO_3$ , along with $H_2O$

Similarly, $Na_2CO_3$ is formed with the reaction of $NaOH$ and $H_2CO_3$ , along with $H_2O$ .

But $Na_2CO_3$ is less soluble, and hence is more easily prepared.