The laboratory technician accidentally mixed NaHCO3 with BaCl2*2H2O instead of KCl when preparing unknown mixtures for this experiment.
How would this mistake affect the calculated percentage of NaHCO3 in the mixture? Justify your answer with an explanation.
Frankly, it makes no sense. If he accidently spilled XXX into something,why would be be calculating the percentage of XXX? It shouldn't be there at all.
I think what you mean is the unknown is supposed to be a mixture of NaHCO3 and KCl but instead it is a mixture of NaHCO3 and BaCl2.2H2O. It won't make any difference if you are determining NaHCO3 by titration with an acid.
This mistake would affect the calculated percentage of NaHCO3 in the mixture because BaCl2*2H2O (barium chloride dihydrate) can react with NaHCO3 (sodium bicarbonate) to form a precipitate. This precipitate is insoluble in water and would cause the loss of NaHCO3 from the mixture.
The balanced chemical equation for the reaction between BaCl2*2H2O and NaHCO3 is:
BaCl2*2H2O + 2NaHCO3 -> Ba(HCO3)2 + 2NaCl + 2H2O
As per the reaction, instead of NaHCO3 being present in the final mixture, Ba(HCO3)2 (barium bicarbonate) would be formed along with NaCl (sodium chloride) and water. This means that the amount of NaHCO3 calculated in the mixture would be lower than the actual amount, resulting in an underestimated percentage.
To obtain an accurate percentage of NaHCO3 in the mixture, it is essential to use the correct chemical (KCl) for the experiment rather than BaCl2*2H2O.
To understand how this mistake would affect the calculated percentage of NaHCO3 in the mixture, we need to consider the chemical reactions that occur when NaHCO3 and BaCl2*2H2O are mixed.
When NaHCO3 (sodium bicarbonate) reacts with BaCl2*2H2O (barium chloride dihydrate), a double displacement reaction occurs:
2NaHCO3 + BaCl2*2H2O -> 2NaCl + Ba(CO3)2 + 4H2O
This reaction results in the formation of sodium chloride (NaCl), barium carbonate (Ba(CO3)2), and water (H2O). In this case, the NaHCO3 is no longer present in the mixture.
As a result, when calculating the percentage of NaHCO3 in the mixture, we would not be able to obtain an accurate result. This is because the NaHCO3 has undergone a chemical reaction, converting into other compounds, and is no longer present in its original form.
Therefore, the mistake of mixing NaHCO3 with BaCl2*2H2O instead of KCl would lead to an incorrect calculation of the percentage of NaHCO3 in the mixture. To obtain an accurate measurement, it would be necessary to start the experiment again with the correct mixture of NaHCO3 and KCl.