Purity factor of the primary standard loo Single sample weight H 84 mg Aliquots

Question

Purity factor of the primary standard loo Single sample weight H 84 mg Aliquots used: mL 25mL 85 mL Titrant used: 25.20L 35 mL2u.OOmL Mof HCI ool13 0.01408 0.01157 M of HCI overall "stats" mean .Oo5A median 0.01757 RSD 13.3pt refined "stats": meanO785 medianRSD 32ppt comments (on the back) RSD 22p DETERMINATION OF PURITY OF UNKNOWN SODA ASH SAMPLE M of HCl used .01785 Single sample weight 873 mg Titer: 1 mL HCl= mg SA #3 3omL 0 mL 50ml 50ml 5 mL 50 mL Aliquots used: L (0.00 mL 11351 RSDpt Titrant used: Soda Ash purity(wt% Na2CO3): l.19 19 1.1729 3.043 10- overall "stats mean Ll ol median L13 median RSDppt refined "stats"mean comments (on the back) Central value being reported for grade: LI7 29 % Na2CO3

Explanation / Answer

The reaction between Na2Co3 and HCl procees via the following reaction:

Na2CO3 + 2HCl -> NaCl + H2O + CO2

Now, when 1mL of HCl is added, let mass of SA required be x.so,

molar mass of SA = 106 g/mol

so number of moles = x/106

hence two moles of HCl requires one mole of SA.

so 1 mL of HCl requires y milli moles amount of SA

where y = 1*0.01785

or 1*0.01785 = 2*(x/106)

or x=0.946 mg of SA

This is the incorrect part in your calculation.

to solve for the total value, the total volume of the sample is needed, but here only the aliquot sample is given.

you will then have to find out the amount of SA required for the titrant and then multiply with the dilution factor. that will be total amount of SA in mg present.

divide the following by 287.3 and then multiply by 100 (to convert the result in percentage)

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