Finding percentage of carbonate in mixed sample titration

[thatgirlyouknow]

[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.

 

[eli64]

Should I assume both dissolve equally? .

answer this question after looking up a solubility table. Only one of the carbonates is in the solution that the acid is titrating. use the mol of carbonate = vol acid x M acid /2 ; then find g of carbonate using MM. you have the mass of the mixture; %= g carbonate/g mixture

 

[chemisttree]

answer this question after looking up a solubility table. Only one of the carbonates is in the solution that the acid is titrating. use the mol of carbonate = vol acid x M acid /2 ; then find g of carbonate using MM. you have the mass of the mixture; %= g carbonate/g mixture

NO.

Instead, find the total number of moles of acid used in the titration. Write out the neutralization equation, ignoring the Sr and Li for the time being. This will be used to determine the total number of moles of carbonate titrated. The total number of moles of carbonate titrated will be equal to the total number of moles of SrCO3 plus LiCO3.

let
x=grams of SrCO3
y=grams of LiCO3

therefore, x + y = 0.5310g (or y = 0.5310 g - x) (1)

and

# moles SrCO3 = x / 147.63
# moles LiCO3 = y / 73.89 (or (0.5310 - x) / 73.89 (2)

and

# moles SrCO3 + # moles LiCO3 = moles carbonate (determined from HCl titration)
rewriting the above using equation (2) yields,

x/147.63 + (0.5310 - x)/73.89 = moles carbonate (determined from HCl titration)

Solve for x and substitute into equation (1) and determine y.

 

[eli64]

sorry, filtration first

Yes, of course the acid would react with both carbonates. I was thinking experimentally and not reading the problem (!). If this was a lab problem could you not dissolve the mixture of carbonates and and filter the insoluble one then titrate the soluble one. The question asked for the percentage of the Li2CO3.

sorry to mislead

 

[physicsuserX]

I am still confused as to how you would go from:

x/147.63 + (0.5310 - x)/73.89 = moles carbonate (determined from HCl titration)

into solving for x.

My question is how do I obtain the moles of carbonate?