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thatgirlyouknow
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[SOLVED] Finding percentage of carbonate in mixed sample titration
A sample of mixed carbonates, SrCO3 (147.63 g/mol) and Li2CO3 (73.89 g/mol), weighed 0.5310 grams. If this sample is titrated with acid, it is found that 29.47 mL of 0.3400 M HCl are required to reach the endpoint. What is the percentage of Li2CO3 in the sample? SrCO3 and Li2CO3 are the only components in the sample and carbonate reacts as a di-protic base.
Ok so I know:
SrCO3 + 2 HCl --> CO2 + H2O...
LiCO3 + 2HCl --> CO2 + H2O...
So the stoichiometric ratio is two for both (hence the diprotic base). However, after setting up:
mol HCl = mol SrCO3 x 2 (SR) + mol LiCO3 x 2 (SR)
and changing the mol into wt over molar mass, where do I go from here? Should I assume both dissolve equally? If so, then the percent Li is 25 (one fourth), but that seems way too easy.. especially since they have different molar masses.
A sample of mixed carbonates, SrCO3 (147.63 g/mol) and Li2CO3 (73.89 g/mol), weighed 0.5310 grams. If this sample is titrated with acid, it is found that 29.47 mL of 0.3400 M HCl are required to reach the endpoint. What is the percentage of Li2CO3 in the sample? SrCO3 and Li2CO3 are the only components in the sample and carbonate reacts as a di-protic base.
Ok so I know:
SrCO3 + 2 HCl --> CO2 + H2O...
LiCO3 + 2HCl --> CO2 + H2O...
So the stoichiometric ratio is two for both (hence the diprotic base). However, after setting up:
mol HCl = mol SrCO3 x 2 (SR) + mol LiCO3 x 2 (SR)
and changing the mol into wt over molar mass, where do I go from here? Should I assume both dissolve equally? If so, then the percent Li is 25 (one fourth), but that seems way too easy.. especially since they have different molar masses.