Final answer:
To determine the mass of sodium bicarbonate added, calculate the moles of NaOH used in the neutralization, use it to find the moles of HNO3 that reacted, and then determine the corresponding moles of NaHCO3. Option D is correct.
Step-by-step explanation:
The question is asking how many grams of sodium bicarbonate (NaHCO3) were added to a 0.25 M solution of HNO3, after it was titrated with 168 mL of 0.15 M NaOH. To find the grams of sodium bicarbonate, we must first determine the moles of NaOH used in the neutralization reaction. Then, we'll use this information to find the moles of HNO3 that reacted, which in turn will give us the moles of NaHCO3 that were neutralized by the HNO3 initially.
Since NaOH reacts with HNO3 in a 1:1 molar ratio:
Calculate moles of NaOH: Moles of NaOH = 0.15 M × 0.168 L = 0.0252 moles.
Since the molar ratio of NaOH to HNO3 is 1:1, the moles of HNO3 that reacted with NaOH is also 0.0252 moles.
Subtract the moles of HNO3 that reacted with NaOH from the initial moles of HNO3 to find moles of HNO3 that reacted with NaHCO3: Initial moles of HNO3 = 0.25 M × 0.125 L = 0.03125 moles; Remaining moles of HNO3: 0.03125 moles - 0.0252 moles = 0.00605 moles.
The moles of HNO3 that reacted with NaHCO3 are the same as the moles of NaHCO3 initially present since they also react in a 1:1 ratio.
Calculate the mass of NaHCO3: Mass = Moles × Molar mass of NaHCO3; Molar mass of NaHCO3 = 84.01 g/mol. Mass of NaHCO3 = 0.00605 moles × 84.01 g/mol = 0.508 g.
Therefore, the mass of sodium bicarbonate added was 0.508 grams.