Final answer:
To find the required volume of 0.1 M HCl to neutralize the Na₂CO₃ and NaHCO₃ mixture, calculate the moles, apply the stoichiometry of the reaction, and divide by the HCl concentration.
Step-by-step explanation:
The volume of 0.1 M HCl required to neutralize completely 4g of an equivalent mixture of Na₂CO₃ and NaHCO₃ can be found by first determining the moles of the sodium carbonate and sodium bicarbonate in the mixture. Next, we use the balanced chemical equations for their neutralization.
For Na₂CO₃, the equation is: Na₂CO₃(aq) + 2 HCl(aq) → 2 NaCl(aq) + H₂O(l) + CO₂(g) and for NaHCO₃, the equation is: NaHCO₃(aq) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g). We calculate the moles of each compound in the mixture, then find the total moles of HCl needed.
Since Na₂CO₃ requires 2 moles of HCl for every mole and NaHCO₃ requires 1 mole of HCl per mole. Finally, we divide the total moles of HCl by the concentration of the HCl solution to find the required volume.