I have no idea what procedure is described in your lab manual. Also, by mentioning s.f. makes me wonder if that is 11 mL or 11.00.
CaCO3 + 2HCl ==> CaCL2 + H2O + CO2
total mols HCl before rxn = M x L = ?
mols HCl left over after rxn = MNaOH x L NaOH = ?
mols HCl used in the rxn = total mols - mols left over.
Convert mols HCl used in the rxn to mols CaCO3 in the rxn by using the coefficients in the balanced equation.
Convert mols CaCO3 to grams. g = mols x molar mass = ?
Round to the correct number of s.f.
A TUMS tablet was dissolved in 100.00 ml of 0.1805 M HCl and analyzed for CaCO3 according to the procedure described in the lab manual. It took 11 ml of 0.1491 M NaOH to reach the endpoint. Calculate the mass of CaCO3 in TUMS tablet to the correct number of significant .
2 answers
Ok, you put initially excess HCL, then back titrated with NaOH to enpoing
backpoint:
NaOH+HCl>>NaCl + H20
moles HCl titrated= molesNaOH
= .1419*.011=.00156
moles HCl left: moles orig-0.00156
= .100*.1805-.00156== 0.01649
and that HCL was used to titrate the original CaCO3
moles CaCO3=1/2 moles HCL used
=1/2 * 0.01649=.00825
grams CaCO3=.00825*100.1=.825 grams
So check my work. Regular tums have about .5grams, other sizes .75, and 1.0 grams, and 1.1 grams.
backpoint:
NaOH+HCl>>NaCl + H20
moles HCl titrated= molesNaOH
= .1419*.011=.00156
moles HCl left: moles orig-0.00156
= .100*.1805-.00156== 0.01649
and that HCL was used to titrate the original CaCO3
moles CaCO3=1/2 moles HCL used
=1/2 * 0.01649=.00825
grams CaCO3=.00825*100.1=.825 grams
So check my work. Regular tums have about .5grams, other sizes .75, and 1.0 grams, and 1.1 grams.