Question

What is happening at the molecular level to explain your observations? In other words, what's a scientific model that's representative of your data? The lab experiment instructs you to react 0.21 g of NaHCO3 with excess CH3COOH. There are different concentrations of NaOH: 0.01 M, 0.001 M, 0.0001 M, and 0.00001 M.

Answer 1

In the reaction between NaHCO3 and CH3COOH, NaCH3CO2, water, and CO2 are formed. When titrating with NaOH, the amount of CH3COOH that reacts is equivalent to the amount of NaOH added, while any excess CH3COOH remains unreacted. Titrations with standard solutions like HCl can be used to find the concentration and mass of other compounds in solution.

Step-by-step explanation:

At the molecular level, when NaHCO3 reacts with CH3COOH, a double displacement reaction occurs, forming NaCH3CO2 (sodium acetate), water (H2O), and carbon dioxide (CO2). When NaOH is added to this system, it reacts with CH3COOH in a 1:1 stoichiometry. This means that for every mole of NaOH added, one mole of CH3COOH will be neutralized, producing one mole of acetate ion (CH3CO2-) and leaving an equivalent amount of excess CH3COOH.

For example, if 5.00 mmol of CH3COOH is initially present and 1.00 mmol of NaOH is added, the reaction of OH- with CH3COOH will leave a final amount of 4.00 mmol CH3CO2H and produce 1.00 mmol of CH3CO2-. Similarly, if 1.0 × 10^-4 mol of NaOH is added to a solution, it will neutralize an equivalent amount of CH3COOH, and the final product will be 1.0 × 10^-4 mol of NaCH3CO2.

Titrations can be used to determine concentrations of solutions and masses of substances involved. For instance, if 25.66 mL of 0.1078 M HCl is used to titrate an unknown sample of NaOH, the number of moles of HCl reacted (0.002766 mol) will equal the number of moles of NaOH, allowing us to calculate the mass of NaOH using its molar mass.