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what volume of 0.110 m hcl is required for the complete neutralization of 1.00 g of nahco3 (sodium bicarbonate)?

User Pflz
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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.

User BlackShadow
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