Author Topic: Chemistry Titration help required (Read 1247 times) Tweet Share

psychlaw · « **on:** September 10, 2008, 10:50:30 pm »

THANKS .... all done now

bucket · « **Reply #1 on:** September 10, 2008, 11:13:35 pm »

b) you use stoichiometry. calculate the amount of mol in 17.5ml of HCl then use that to find out the amount of mol of NH4 using the equation.
c)i don't understand why you diluted the solution since it doesn't really do anything in this case. the amount of mol in the original sample should be the same as the amount of mol in the diluted solution since you haven't taken any aliquots from it. use that mol to find the mass $m=n\times M$ then divide that answer by the mass in the original sample (which is $41.2g \times 0.02L$ ) and multiply that by 100 to get the % by mass.

bucket · « **Reply #2 on:** September 10, 2008, 11:19:19 pm »

i'm not 100% sure on the % by mass part because I don't know whether they meant mass/mass or mass/volume. ask mao about that.

psychlaw · « **Reply #3 on:** September 10, 2008, 11:22:38 pm »

thanks, I got B, 0.01925 mol, I think its right

C is confusing me aswell =/
its something to do with the 250/20 though

bucket · « **Reply #4 on:** September 10, 2008, 11:25:34 pm »

the 250/20 would only apply if you diluted the original solution to 250ml then taken 20mls out of that diluted solution and titrated it against the HCL. This is because there would be 12.5 (250/20) the amount of ammonia in the original sample than in the 20ml aliquot if you did do that(only using 20ml out of the 250ml that you had originally, so there is less ammonia.

from what you've told me though, you didn't do that.

psychlaw · « **Reply #5 on:** September 10, 2008, 11:26:59 pm »

Quote from: bucket on September 10, 2008, 11:25:34 pm

the 250/20 would only apply if you diluted the original solution to 250ml then taken 20mls out of that diluted solution and titrated it against the HCL. This is because there would be 12.5 (250/20) the amount of ammonia in the original sample than in the 20ml aliquot if you did do that(only using 20ml out of the 250ml that you had originally, so there is less ammonia.

from what you've told me though, you didn't do that.

oh shit my bad, poor explaining

...

we did actually dilute the original solution to make it 250 mL

bucket · « **Reply #6 on:** September 10, 2008, 11:28:03 pm »

did you then take a fraction of that diluted solution out to titrate against hcl?

bucket · « **Reply #7 on:** September 10, 2008, 11:35:40 pm »

is this how it went?

psychlaw · « **Reply #8 on:** September 10, 2008, 11:36:44 pm »

hmmm I got 0.01925 mol

how do I find the mass if ammonia in the orginal sample $:-\$

EDIT: THANKS HEAPS BUCKET, got it now! thanks!

bucket · « **Reply #9 on:** September 11, 2008, 12:02:22 am »

$ n(NH_4OH)_{diluted}= 0.11 \times 0.0175 = 0.001925mol $

$ n(NH_4OH)_{initial}= 0.001925 \times \frac{250}{20} = 0.02406mol $

$ 0.02406mol \times 35$ (molar mass) $= 0.8422g/20ml \times 50$ (to make in litres) $= 42.12g/L $

psychlaw · « **Reply #10 on:** September 11, 2008, 12:03:08 am »

Quote from: bucket on September 11, 2008, 12:02:22 am

$ n(NH_4)_{diluted}= 0.11 \times 0.0175 = 0.001924mol $

$ n(NH_4)_{initial}= 0.001925 \times \frac{250}{20} = 0.02406mol $

$ 0.02406 \times 35 = 0.8422g/20ml \times 50$ (to make in litres) $= 42.12g/L $

dude you are a legend!

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Author Topic: Chemistry Titration help required (Read 1247 times) Tweet Share

psychlaw

Chemistry Titration help required

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