Question

0.25 mole of hydrogen chloride was dissolved in distilled water and the volume made up of 0.50 dm3. If 15.0 cm3 of the solution requires 12.50 cm3 of aqueous sodium trioxocarbonate (IV) for neutralization, calculate the concentration of the alkaline solution
A. 0.30 mol dm3
B. 0.40 mol dm3
C. 0.50 mol dm3
D. 0.60 mol dm3

Please show working....

Answer 1

Na₂CO₃ + 2HCl -->2NaCl + H₂O+CO₂;

The quantity of HCl that reacted is: n(HCl)=15*0.25/50=0.075 mol

The quantity of Na₂CO₃ that reacted is: n(Na₂CO₃)=1/2*n(HCl)=0.0375 mol;

C(Na₂CO₃)=n(Na2CO3)/V(solution)=0.0375/0.0125= 3 mol/dm³