Question

B. if 6.73 g of na2co3 is dissolved in enough water to make 250. ml of solution, what is the molar concentration of sodium carbonate? what are the molar concentrations of na+ and co32- ions? (note 1 mol na2co3 = 2 mol of na+ = 1 mol co32-)

Answer 1

Answer:- $[Na_2CO_3]=0.254M$ , $[Na^+]=0.508 M$ , $[CO_3^2^-]=0.254M$

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL.  Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

$\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})$

= $0.254MNa_2CO_3$

So, molarity of sodium carbonate solution is 0.254 M.

sodium carbonate dissociate to give the ions as:

$Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-$

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.

$[Na^+]=2(0.254M)$ = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

$[CO_3^2^-]=0.254M$