200cm3 of saturated Na2CO3 solution was reacted with 100cm3 of 2 mol/dm3 HCL solution in a beaker at 45°C . Calculate the solubility of Na2CO3 solution in mol/dm3
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
moles HCl used = M x dm3 = 2 mol/dm3 x 0.100 dm3 = 0.2 mole
moles Na2CO3 neutralized = 1/2 x 0.2 = 0.1 mol Na2CO3 in that 200 cc saturated solution. (Na2CO3) = 0.1 mol/0.2 dm3 = 0.5 mol/dm3 @ 45 C.
Personally, I think it is cumbersome to change back and forth from cc to dm3 to L. Also I think it so much easier to use molarity = M = mols/L instead of mols/dm^3. BUT, I'm not teaching the class.