Answer Happy

A 15.00 g sample containing mixed alkali and other inert
components was dissolved and diluted to 300 mL with water. A 20 mL
aliquot was titrated with 5.02 mL of 0.5352 M HCl to reach PHP
endpoint. Another 20 mL aliquot was titrated to the BCG endpoint,
using up 18.87 mL of titrant in the process. Note: Answer in two
decimal places only. If there is no answer, type in 0.00. Use the
molar masses 105.989 𝑔/𝑚𝑜l 𝑁𝑎2cO3 and 84.007 g/mol NaHCO3
indicated. The volume of titrant needed to neutralize NaOH is
_________ mL. The volume of titrant needed to neutralize Na2CO3 is
_____________ mL. The volume of titrant needed to neutralize NaHCO3
is _____________ mL. The mass of NaOH is ___________ g. The mass of
Na2CO3 is __________ g. The mass of NaHCO3 is __________ g. The
percent weight of NaOH is __________ %. The percent weight of
Na2CO3 is __________ %. The percent weight of NaHCO3 is __________
%.
The choices for percent weight are:
a. 0
b. 10.75
c. 17.56
d. 28.48
e. 39.70
6. A 10.00 g sample containing mixed alkali and other inert components was dissolved and diluted to 500 ml with water. A 25 mL aliquot was titrated with 14.35 mL of 0.1202 M HCl to reach PHP endpoint. Another 25 mL aliquot was titrated to the BCG endpoint, using up 38.50 mL of titrant in the process. Identify the alkali components of the sample and the % weight of each. = This is a double titration of the alkali mixture problem. In determining the mixture component, we compare the volume of titrant used to reach PHP and BCG endpoint. Vphp = 14.35 mL and Vecg = 38.50 mL or Vecg = 19.25 mL. Since Vphp < Vecg, we can infer that this mixture is composed of Na2CO3 and NaHCO3. Using the similar table in the previous item, we track the volumes needed to neutralize the species. 2 = BCG 9.8 ml NaHCO3 end 14.35 ml 14.35 ml Na2CO3 NaHCO3 end Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2 NaHCO3 + HCl → NaCl + H2O + CO2 Vphp/VEP1 14.35 mL Vep2 24.15 ml VBCG 38.5 mL From the diagram, 9.8 mL of HCl is needed to neutralize NaHCO3 and 28.7 mL is needed to completely neutralize Na2CO3. Using the mole relationship from the chemical equation written above, we can solve for the mass of the components and the percent weight.