Question

Baking soda (NaHCO3) can be added to a fruit mix solution to create a carbonated drink. An example is the reaction between baking soda and citric acid below.
C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3H2O + 3CO2

B. How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda? Show your work.

Answer 1

Answer:

74.4 ml

Explanation:

          C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)

Given     15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate

From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.

Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.

The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution

=> Volume (Liters) = moles citric acid / Molarity of citric acid solution

=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml