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Old September 24th, 2006, 13:02
_cheers _cheers is offline
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Default HELP! PLS! ASAP! Back titrations

Ok, I love Chem, but I cannot get these back titrations! I have a lab question and I'm really stuck.

Its determining the molar mass of an unknown metal carbonate: 1.85 g

this is diluted with distilled water up to 250 ml, then 20 ml of solution is mixed with 15 ml of .1789M HCL. This is then back titrated with 17.56 ml .1013M NAOH.

I kow I need to find the acid added, the acid unused to find the acid used, but I keep getting the wrong numbers. I'm not sure what to make of the dilution into 250 ml or where to add it into the equation...

HELP! :shock:
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Old September 24th, 2006, 16:51
_cheers _cheers is offline
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Default question: continued

Ok this is what I've done:

rx 1) M2CO3 +2HCL ->2MCL + CO2 + H20
rx 2) HCL + NaOH -> NaCl + H20

Mole acid added: (.015L)(.1789M)= .00268 mole HCL in 20 ml of solution
mole acid in excess (.1013M) (.01756L) NaOH = .001778 mole HCL
.00268-.001778 = .0009 mole HCL reacted with M2CO3 in 20 ml

moles of M2CO3 = .5(.0009) = .000452 mol M2CO3

mm= 1.85g/ .000452 = 4093g (wrong!)

2 x M +12+48 = 4093
2M = 4093/2
= 2016.5

obviously wrong...but where??? I have the exact same question with different numbers that works with this procedure...I dont understand!

thanks
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Old September 25th, 2006, 13:51
adrian adrian is offline
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Default back titration

The reason is the presence of carbonate. As you observed, CO2 is generate during the "direct" titration.
Because of that, is difficult to identify precisely the equivalence point (partial solubilised into the water). Second, carbonic acid is weak and for precise dosage would be hard to choose the appropriate mixture of pH indicators.
Then, this method supersedes all the inconvenience using a titration of a hard acid (hydrocloric) with a hard base.
If you really love Chem, you'll easelyfind some theoretic description of acid-base titration curves. You'll find useful.
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