# Question #8ce57

Mar 30, 2015

The concentration is 0.17M

Sodium carbonate dissociates:

$N {a}_{2} C {O}_{3 \left(s\right)} + \left(a q\right) \rightarrow 2 N {a}_{\left(a q\right)}^{+} + C {O}_{3 \left(a q\right)}^{2 -}$

Since 1 mole sodium carbonate gives 2 moles Na+ its concentration must be half i.e 0.34/2 = $\text{0.17M}$

Mar 30, 2015

The concentration of sodium carbonate will be $\text{0.17 M}$.

The quickest way to determine the concentration of the sodium carbonate solution is to have a look at its dissociation in aqueous solution

$N {a}_{2} C {O}_{3} \to \textcolor{b l u e}{2} N {a}^{+} + C {O}_{3}^{2 -}$

Notice that 1 mole of sodium carbonate produces $\textcolor{b l u e}{2}$ moles of sodium cations and 1 mole of carbonate anions.

Molarity is defined as moles of solute per liter of solution, which means that, in a given volume, you'll have twice as many moles of sodium cations than of sodium carbonate.

Automatically, the molarity of the sodium cations will be twice that of the sodium carbonate.

A a result, the molarity of $N {a}_{2} C {O}_{3}$ will be half that of the $N {a}^{+}$

${C}_{N {a}_{2} C {O}_{3}} = {C}_{N {a}^{+}} / 2 = \textcolor{g r e e n}{\text{0.17 M}}$