How many mg of NaHCO3 are in a 500-mg tablet if...?

if 40 ml of 0.120 M HCl is required to neutralize this sample? can you explain?

Answer:
The answer is derived from being able to balance the equation using the fact that 40 mL (or 0.040 liters) of acid at 0.120 M (moles/liter) neutralizes the base. That's 0.0048 moles. The molecular weight of sodium bicarbonate is 23 (Na) + 61 (HCO3) or 84. 84 grams/mole times 0.0048 moles gives the amount of sodium bicarbonate (in grams). Convert to mg for the final answer.

Always remember to carry units with the numbers and the math becomes much easier.

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