Question

Question #8ce57

Answer 1

The concentration is 0.17M

Sodium carbonate dissociates:

`Na_2CO_(3(s))+(aq)rarr2Na_((aq))^(+)+CO_(3(aq))^(2-)`

Since 1 mole sodium carbonate gives 2 moles Na+ its concentration must be half i.e 0.34/2 = `"0.17M"`

Answer 2

The concentration of sodium carbonate will be `"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

`Na_2CO_3 -> color(blue)(2)Na^(+) + CO_3^(2-)`

Notice that 1 mole of sodium carbonate produces `color(blue)(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 `Na_2CO_3` will be half that of the `Na^(+)`

`C_(Na_2CO_3) = C_(Na^(+))/2 = color(green)("0.17 M")`