Question #8ce57

2 Answers
Mar 30, 2015

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"#

Mar 30, 2015

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")#