If 5 g of sodium bicarbonate is combined with water to make a 100 ml solution, how do I find out how many grams in the final solution?

Question

In chem lab we did a quality control test to determine if the label for sodium bicarbonate (baking soda) is accurate. The label says there is 1,258.56 mg of sodium in a 5 g serving.

We made a solution using 5 g of sodium bicarbonate and 100ml of DI water. We titrated 20ml of the solution with .1M HCl standardized solution, and the average amount of HCl used after 3 titrations was 22.23 ml to reach end point.

 

 Given this information, how do I compare our results to determine how close they are to the advertised value?

 

I am having a hard time figuring out grams, as I am not 100% sure what our concentration is. I am not sure even if the "sodium" listed would be NaCl, the product, or the sodium bicarbonate itself, NaHCO3. I am assuming this will have an effect on the results.

I have done M1V1 = M2V2 and get .1115 moles of sodium bicarbonate used but not sure if it is accurate, as that is 9.335g, which is a lot higher than the 1.26 g of sodium per 5 g that is listed on the label.

 

Can someone please help me with steps to compare the results to the advertised values on the label?

PLEASE HELP if you can!! :)

Answer 1

The mass of NaHCO₃ in each aliquot is 5 g (20 mL / 100 mL) = 1 g.

For this experiment each mole of NaHCO₃ requires 1 mole of HCl to reach the equivalence point.  Also, there is 1 mole of Na⁺ in each mole of NaHCO₃.

To solve the problem, forget about mL₁ x M₁ = mL₂ x M₂.  Just use moles.

moles HCl = 22.23 mL (1 L / 1000 mL)(0.1 mol HCl / L) = 0.002223 moles

moles Na⁺ = 0.002223 mol HCl (1 mol Na⁺ / mol HCl) = 0.002223

g Na⁺ = 0.002223 mol Na⁺ (22.9898 g Na⁺) / (mol Na⁺) = 0.051106 g = 51.106 mg Na⁺

To put this on a 5 g sample basis, multiply by 5:  51.105(5) = 255.53 mg / 5 g sample.

Your label lists the Na⁺ value as 1258.56 mg / 5 g sample, so something appears to be wrong.  For 100% pure NaHCO₃, the level of Na⁺ should be 5000(22.9898)/(84.0066) = 1368.3 mg / 5 g sample.  

 

By the way if you really mixed 5 g of sample with 100 mL of water (rather than making 100 mL of solution using a 100 mL volumetric flask), the aliquot factor will not be known very accurately since 5 g sample / 100 mL water is not the same as 5 g sample / 100 mL solution