Calculate the mass of sodium bicarbonate needed to neutralize 3.0 mL of 3.0 M HCl. Show your work.
To calculate the mass of sodium bicarbonate needed to neutralize the given volume and concentration of hydrochloric acid (HCl), we need to follow a series of steps:
Step 1: Write down the balanced chemical equation for the neutralization reaction between HCl and sodium bicarbonate (NaHCO3). It is as follows:
HCl + NaHCO3 --> NaCl + H2O + CO2
Step 2: Determine the molarity (M) of the acid solution and the volume (V) used. In this case, we are given that the HCl has a concentration of 3.0 M and a volume of 3.0 mL. However, to use these values in calculations, we need to convert the volume to liters. Therefore:
Volume (V) = 3.0 mL = 3.0 mL × (1 L / 1000 mL) = 0.003 L
Step 3: Apply the concept of molarity (M) to calculate the number of moles (n) of HCl present in the given volume. The formula is:
n = M × V
n = 3.0 M × 0.003 L = 0.009 mol
Step 4: Utilize the stoichiometry of the balanced equation to determine the mole ratio between HCl and NaHCO3. The balanced equation tells us that 1 mole of HCl reacts with 1 mole of NaHCO3. Therefore, we can conclude that 0.009 moles of HCl will react with 0.009 moles of NaHCO3.
Step 5: Identify the molar mass of sodium bicarbonate (NaHCO3). It consists of one sodium (Na) atom with a molar mass of 22.99 g/mol, one hydrogen (H) atom with a molar mass of 1.01 g/mol, one carbon (C) atom with a molar mass of 12.01 g/mol, and three oxygen (O) atoms with a combined molar mass of 47.99 g/mol:
Molar mass (NaHCO3) = (1 × 22.99) + (1 × 1.01) + (1 × 12.01) + (3 × 16.00) = 84.01 g/mol
Step 6: Convert moles of NaHCO3 to grams using the molar mass of NaHCO3 and the number of moles calculated earlier:
Mass = moles × molar mass
Mass = 0.009 mol × 84.01 g/mol = 0.756 g
Therefore, you would need approximately 0.756 grams of sodium bicarbonate to neutralize 3.0 mL of 3.0 M HCl.