3) 2.413 g sample contains Na CO,, NaHCO, and inert material. This sample was solved in water and diluted to 250.0 ml. 80 ml solution was taken and titrated with 0.09644 N HCl with the indicator of phenolftalein. 6.13 ml HCl was consumed. From a second solution, again 80 ml was taken and titrated with the same HCl solution with the indicator of methylorange. 15.4 ml HCl was consumed. Calculate the percentage of NaHCO, and Na CO in sample (Na CO=106.0 g/mol, NaHCO =84.02 g/mol).
3) 2.413 g sample contains Na CO,, NaHCO, and inert material. This sample was solved in water and diluted to 250.0 ml. 80 ml solution was taken and titrated with 0.09644 N HCl with the indicator of phenolftalein. 6.13 ml HCl was consumed. From a second solution, again 80 ml was taken and titrated with the same HCl solution with the indicator of methylorange. 15.4 ml HCl was consumed. Calculate the percentage of NaHCO, and Na CO in sample (Na CO=106.0 g/mol, NaHCO =84.02 g/mol).
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Transcribed Image Text:3) 2.413 g sample contains Na CO,, NaHCO, and inert material. This sample was solved
in water and diluted to 250.0 ml. 80 ml solution was taken and titrated with 0.09644 N
HCl with the indicator of phenolftalein. 6.13 ml HCl was consumed. From a second
solution, again 80 ml was taken and titrated with the same HCl solution with the indicator
of methylorange. 15.4 ml HCl was consumed. Calculate the percentage of NaHCO, and
Na CO in sample (Na CO=106.0 g/mol, NaHCO =84.02 g/mol).
Expert Solution
Step 1
Following are the two reactions and their indicators can be represented as:
Sample contain
Na2CO3 + NaHCO3 + Inert Material = 2.413 g in 250 mL water.
Reaction 1 for Na2CO3:
Step 2
Now calculate the number of moles of sodium carbonate in 250 mL sample solution:
Now calculate weight of sodium carbonate in the sample as:
Step 3
Now percentage of Na2CO3 can be calculated as:
Reaction 2 for combine Na2CO3 and NaHCO3
Note: first Na2CO3 convert into NaHCO3, therefore next there will be 3 times of the NaHCO3, two from Na2CO3 and 1 from NaHCO3
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