Answer:
approximately 164 mL of 0.110 M HCl is required for the complete neutralization of 1.00 g of NaHCO3.
Step-by-step explanation:
To determine the volume of 0.110 M HCl (hydrochloric acid) required for the complete neutralization of 1.00 g of NaHCO3 (sodium bicarbonate), we need to consider the stoichiometry of the reaction between HCl and NaHCO3.
The balanced chemical equation for the reaction is:
NaHCO3 + HCl -> NaCl + H2O + CO2
From the equation, we can see that the molar ratio between NaHCO3 and HCl is 1:1. This means that 1 mole of NaHCO3 reacts with 1 mole of HCl.
First, let's calculate the number of moles of NaHCO3:
moles of NaHCO3 = mass / molar mass
moles of NaHCO3 = 1.00 g / (22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol))
moles of NaHCO3 ≈ 0.018 mol
Since the reaction is 1:1, we need the same number of moles of HCl. Now we can calculate the volume of 0.110 M HCl needed:
moles of HCl = 0.018 mol
volume of HCl = moles of HCl / molarity
volume of HCl = 0.018 mol / 0.110 mol/L
volume of HCl ≈ 0.164 L or 164 mL
Therefore, approximately 164 mL of 0.110 M HCl is required for the complete neutralization of 1.00 g of NaHCO3.