Mixture of 1 g each of Na 2 CO 3 and NaHCO 3 is reacted with 0.1 N HCl. The quantity of 0.1 M HCl required to react completely with the above mixture is

Step-by-step solution

Na 2 CO 3 + 2 HCl → 2 NaCl + H 2 O + CO 2 1 mol 2 mol Na 2 CO 3 = 1 g = 1 106 m o l . N a 2 C o 3 ≡ 2 106 m o l . H C l NaHCO 3 + HCl → NaCl + H 2 O + CO 2 NaHCO 3 = 1 g = 1 8 4 mol ≡ 1 8 4 mol HCl Total HCl required = 1 5 3 + 1 8 4 mol HCl Let volume of 0.1 N HCl = V mL then, Moles = 0.1 × V 1 0 0 0 = V 1 0 0 0 0 ∴ V 1 0 0 00 = 1 5 3 + 1 8 4 = 0.0308 mol V = 0.0308 × 1 0 0 0 0 = 308 mL