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### Topic: EDTA Titrations  (Read 434 times)

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#### ubcdumbass

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##### EDTA Titrations
« on: February 14, 2019, 01:54:56 AM »
Question 8

A 0.3284 g sample of brass (a mix of lead, zinc, copper, and tin) was dissolved in nitric acid. The insoluble SnO2•4H2O (s) was removed by filtration. The combined filtrate and washings were diluted to 500.0 mL. This solution was used as a stock solution.

A 10.00 mL aliquot of the stock solution was suitably buffered and titrated for lead, zinc, and copper using EDTA. A volume of 37.56 mL of 2.500 mM EDTA was required to reach the endpoint.

Separately, a 25.00 mL aliquot of the stock was treated with thiosulfate to mask copper, then the lead and zinc were titrated with EDTA. A volume of 27.67 mL of 2.500 mM EDTA was required to reach the endpoint.

Finally, a 100.0 mL aliquot of the stock was treated with cyanide to mask copper and zinc, then the lead was titrated with EDTA. A volume of 10.80 mL of 2.500 mM EDTA was required to reach the endpoint.

Determine the percentage (by mass) of tin in the sample.

This is my work:
https://imgur.com/a/P023M5A

Unsure why i am wrong. any direction in the right way is greatly appreciated.

#### Borek

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##### Re: EDTA Titrations
« Reply #1 on: February 14, 2019, 04:04:13 AM »
I haven't checked the math thoroughly, but the logic looks sound.  Why do you think it is wrong?

The only thing that I am not sure about is the number of sigfigs in the final answer, you are subtracting number that differ by order of magnitude, so the final result has less figures than it appears to have. But that would be heavy nitpicking, as in the real world you should give the answer with information about the error.
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#### ubcdumbass

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##### Re: EDTA Titrations
« Reply #2 on: February 14, 2019, 09:34:02 AM »
Canvas (a site my school uses for homework) said I was wrong

#### chenbeier

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##### Re: EDTA Titrations
« Reply #3 on: February 14, 2019, 10:04:09 AM »
You have to get the moles from each for 500 ml. I think you have forgotten it in the calculation.

For example for the lead you took 100 ml aliquot. The titration gave 0,027 mmol. But for 500 ml it is 5 time more means 0,135 mmol.

The same you have to do for the other two titrations.

My result is 2,65%. Sn

« Last Edit: February 14, 2019, 11:12:18 AM by chenbeier »

#### Borek

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##### Re: EDTA Titrations
« Reply #4 on: February 14, 2019, 04:42:09 PM »
You have to get the moles from each for 500 ml. I think you have forgotten it in the calculation.

As far as I can see OP did it below the line "500 mL = 1/2 L", where masses are calculated using correct volume and concentrations.

Quote
My result is 2,65%. Sn

Final result strongly depends on the rounding errors, so it is basically the same answer.

Mass of tin is 0.3284-0.3200... - so it can't be expressed better than 0.0084 (two significant digits). That means 2.6%. Not that I agree with this approach.
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