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Help me with this Back Titration problem?
A student wishes to analyze an unknown sample, which contains both calcium sulfate and calcium carbonate. The student weighs a 5.000-g sample of the unknown and treats it with 30.0 mL of 1.00 N hydrochloric acid, an amount that was in excess. The excess acid was titrated with 0.155 N NaOH (aq), requiring 25.55 mL to reach equivalence. Determine the percent by weight CaCO3 in the sample.
How does one go about solving this problem? I know that it involves finding the equivalence.
I've been trying to solve it for decades now, decades! Help me, please.
That's actually the right answer... 26%
1 Answer
- hcbiochemLv 71 decade agoFavorite Answer
First, I HATE using normality as a measure of concentration. Molarity just makes so much more sense because it is independent of the specific reaction involved. I'll take a shot.
Moles of HCl initially present = 0.030 L X 1.00 N = 0.0300 mol
Moles HCl remaining = 0.155 X 0.02555 = 3.96 X 10^-3 mol HCl
So 0.0300 - 3.96 X 10^-3 = 0.0260 mol HCl consumed
0.0260 mol HCl ( 1 mol CaCO3/2 mol HCl) = 0.0130 mol CaCO3 X 100 g/mol = 1.30 g CaCO3
1.30 g / 5.00 g X 100 = 26.0% Calcium Carbonate by weight.
