You didn't provide the volume or M of the HCl and you must have that to answer this question.
You added an excess of HCl. How much HCl did you add? moles HCl = M x L = ?
Then you back titrated with NaOH. mols NaOH = M x L = ?
The difference in moles = moles of acid consumed by the TUMS. I would calculate each titration separately and divide the final result of each by the mass of the original sample which should give you moles H^+/grams and that times 2 will give moles/2 grams.
i measured out 2 different tared mass of TUMS sample. One is 0.1554g and the other 0.1546g.
Then i did two titrations with the conclusion of one being 17.91mL of NaOH and the other 17.47mL
my question is: "if you consumed two 1.00g tablets of TUMS, how many moles of acid would be neutralized"??
this is a back titration of standardized HCl with Sodium Hydroxide solution
5 answers
the volume and M of HCl is 0.1312M and 150mL
how do i find then excess of HCl
how do i find then excess of HCl
What volume of 0.1125 M K2Cr2O7 would be required to oxidize 48.16 mL of 0.1006 M Na2SO3 in acidic solution? The products include Cr3+ and SO42- ions.
FDAS
fgbkwbfkw