How many grams of NaHCO3 are needed to make 25.0 mL of a 5.4 M sodium bicarbonate (NaHCO3) solution?

Answer 1

The easiest way for me to do these types of problems was always to balance out the labels (Grams, liters, moles etc.) on everything and work backwards towards your desired answer. For example the problem gives you the Molarity (moles/liter) and asks how many grams are required. This means you need to find a way to convert moles to grams.

So... You're starting with 5.4 M or 5.4 moles/liter of a solution. You need to look at the periodic table of elements and determine that the molecular weight of NaHCO3 would be:

22.99 (1Na) + 1.008 (1H) + 12.01 (1C) + 48 (3O) = 84 grams/mole. To get 1 mole of NAHCO3 exactly, you need to measure out 84 grams.

But you want there to be 5.4 moles/liter, so multiply

5.4 moles/liter X 84 grams/mole = 453.6 grams/liter.

However, the problem does not want an entire liter, rather 25 mL, or 25/1,0000 L = 0.025 L. Therefore,

453.6 grams/liter X 0.025 Liters = 11.34 grams of NaHCO3/0.25 mL