[_cheers]

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...

*delete me* 
 
 
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
 

[Borek]

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

0.000452 mole per 20 mL, not per 250 mL (whole sample).

[_cheers]

ok, thanks, but where did it go wrong? my number is way too large..its supposed to be 326 g/mol

[Borek]

If there is 0.000452 mole in 20 mL, how many moles are in 250 mL of the solution?

[_cheers]

whooooeeeee ) of course...so, its an MCO3, not a M2...so simple when you know how. Thanks SO much! I could just kiss you!  I kiss the screen instead 

Cheers!!!

[Borek]

whooooeeeee ) of course...so, its an MCO3, not a M2...so simple when you know how. Thanks SO much! I could just kiss you!  I kiss the screen instead 

Hold back that kiss for a moment - you are wrong! It is M₂CO₃! And your titration results are pretty precise when it comes to recognition of the cation in the carbonate.

Calculate number of moles in 250 mL and use this information (together with the mass sample) to calculate molar mass of the carbonate.

[_cheers]

ok, i dont think I'm getting this at all 

so, you saying, if I :

.000452 mol x 250 ml/20 ml= .00565 mol m2co3

and take 1.85g/.00565 mol = 327 g/mol (should be 326-that what threw me off)
but if I continue with the above scenario, I end up with a mm of 133.5

Am I making this harder than it should be? 

[Yggdrasil]

1.85 g this is diluted with distilled water up to 250 ml, then 20 ml of solution...

When you calculate the molar mass, you don't divide 1.85g by the number of moles.  You only titrate part of the 1.85g of your compound.  So, before you figure out the molar mass, you first have to find out how many grams of your metal carbonate are in the 20mL of solution you titrated.

[_cheers]

ok, i'll give that a try tommorrow. I have a calc exam at 8 am. Thanks for the help, I'll let you know if it works...if not...oy

cheers

[sdekivit]

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.

the clue here is that you add too much acid and that bit that couldn't react is determined by titrating it.

So here we first have the reaction CO3^2- + 2H₃O⁺ --> 3H2O + CO2

Then all the HCl that didn't reacti is determined by: H₃O⁺ + OH^- --> 2H₂O

To titrate 20 mL of sample 15 mL 0,1789 M HCl ( = 2.6835 mmol H⁺). And this 35 mL is titrated with 17.56 mL 0,1013 M NaOH ( = 1.7788 mmol OH^-)

--> in other words: 1.7788 mmol H⁺ didn't react, so 2.6835 - 1.7788 = 0.9047 mmol H⁺ has reacted according to reaction 1.

Thus in the 20 mL with unknown concentration of carbonate is: (0.9047 / 2) / 20 = 0.02262 M and thus in 250 mL with the same concentration 0.02262 * 250 = 5.654 mmol carbonate was present and thus 5.654 mmol metalcarbonate.

--> molar mass = 1.85 / 5.654 x 10^-3 = 327.2 g/mol

[Borek]

ok, i dont think I'm getting this at all 

IMHO you did great.

.000452 mol x 250 ml/20 ml= .00565 mol m2co3

and take 1.85g/.00565 mol = 327 g/mol (should be 326-that what threw me off)

Remember - this is an experiment, your results are not error free. But it looks like 327 vs 326 is 0.3% wrong - that's pretty good!

but if I continue with the above scenario, I end up with a mm of 133.5

Perfect! Check caesium.

Am I making this harder than it should be? 

You forgot about experimental error and dilution, but otherwise you did OK. Not bad for a first question posted at chemicalforums

[_cheers]

Thanks guys. I couldnt have done it without you! Now I'll kiss the screen