[shylight200]

help please. i have a test tomorrow and i have no idea how to solve this problem


a 650 mg Tums is dissovled in 150 mL of 0.10 M HCl. The excess HCl is titrated to equivalence with 50 mL of 0.1 M NaOH. what is the percentage of CaCO3 in the tums?


CaCO3 + 2HCl ---> CaCl2 + H2CO3
HCl+NaOH---> H2O+NaCl

Tums= 650 mg
CaCO3=?


HCl->
Initial= 0.015 moles HCl
Final= my teacher wrote 0.015-0.05 = 0.010 moles
            i don't know where she got 0.05 from

i don't know how to solve this problem

the answer is 500 mg CaCO3

500/650 * 100 = 77%

[Maz]

So here is the basic idea:

"titrated to equivalence" means that by adding 50ml of .1M NaOH, the excess HCl was neutralized. 

This is a fancy chemist way of saying that adding that amount of base canceled the excess acid, or that Mols of base = Mols of excess acid. 

so (50mL)(.1M NaOH) = 5mmol (milli mol) of OH = mols of excess HCl. 

You know that the TOTAL mols of HCl = (150mL)(.1M HCl) = 15mmol.

Therefore the amount of reacted HCl = Total - excess = 15mmol - 5mmol = 10mmol.

And you can read off from the chemical equation that for every 1 mol of CaCO3 reacted, 2 mol of HCl are needed
or say it backwards,
for every mol of HCl, .5mol of CaCO3 is needed

So mols of CaCO3 reacted = (mol HCl)/2 = 10mmol/2 =  5mmol.

MW of CaCO3 = 100.09g/mol

therefore the Mass CaCO3 reacted = (100.09)(5mmol) = 500mg

And there it is.  Don't let yourself get confused by all the excess information chemists like to throw into their problems.  Just keep your eye on the prize and follow it's path. 

Good luck on your exam.