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    I just joined this site because I was having trouble with some questions I've been given on titration..

    we titrated an unknown acid (25cm3 of 1moldm-3 - which we then diluted with 225cm3 of de-ionised water) with sodium hydroxide (25cm3 of 0.100 moldm-3).

    the question is:

    It was, separately, found that the concentration of the original acid solution was 1.58g of the acid in a volume of 25cm3. Calculate the concentration of the acid in g dm-3 and, hence, deduce the Mr of the acid.

    Please help, I don't even know where to start, I'm so lost with this!! :confused:

    Thanks.
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    ok so start with your acid. You diluted it from 25 cm3 to 225 cm3, so by what ratio has it been diluted? Apply that ratio to your original concentration (1 mol dm-3)
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    Well this is bit of a long problem. You begin my calculating the number of moles in the 25cm3 of your 1mol/dm3 of acid.

    Using Moles=vol x conc you'll get 25/1000 x 1= 0.025moles. Now that you have your moles you find the Mr of the acid by using the formula moles= mass/Mr

    Using this you'll get 1.58/0.025= 63.2 (an ugly figure i know). Now to find the concentration in g/dm3 you use the formula g/dm3= Mr x conc

    now i dont know which concentration you've to use here. If

    1. You use the 25cm3 and your conc is 1mol/dm3 you'll get g/dm3= 63.2 x 1 = 63.2 (i doubt if thats the right answer)

    2. You use the 225 cm3 and your conc will be approx 0.1mol/dm3 (cause you're diluting almost 10 times, ive assumed the volume to be 250cm3 and so my conc is 0.1 but if you do the proper calculation its gonna be slightly less but close to 0.1mol/dm3) In this case your conc in g/dm3 will be = 0.1 x 63.2 = 6.32 (which i think may be the close to the right answer)

    PS: if ive done your homework correctly you owe me quite alot of rep
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    (Original post by bluemax)
    Well this is bit of a long problem. You begin my calculating the number of moles in the 25cm3 of your 1mol/dm3 of acid.

    Using Moles=vol x conc you'll get 25/1000 x 1= 0.025moles. Now that you have your moles you find the Mr of the acid by using the formula moles= mass/Mr

    Using this you'll get 1.58/0.025= 63.2 (an ugly figure i know). Now to find the concentration in g/dm3 you use the formula g/dm3= Mr x conc

    now i dont know which concentration you've to use here. If

    1. You use the 25cm3 and your conc is 1mol/dm3 you'll get g/dm3= 63.2 x 1 = 63.2 (i doubt if thats the right answer)

    2. You use the 225 cm3 and your conc will be approx 0.1mol/dm3 (cause you're diluting almost 10 times, ive assumed the volume to be 250cm3 and so my conc is 0.1 but if you do the proper calculation its gonna be slightly less but close to 0.1mol/dm3) In this case your conc in g/dm3 will be = 0.1 x 63.2 = 6.32 (which i think may be the close to the right answer)

    PS: if ive done your homework correctly you owe me quite alot of rep
    Thank you SO SO much for your help I will rep you first thing in the morning!! It made so much more sense going through it methodically like that, I don't think I would have found that on my own, though. So THANK YOU!!

    I don't suppose you're any good with dot and cross diagrams? It's the last question and I'm about to bang my head into a wall. It's for Sodium Nitrate I just can't work it out (sorry if i sound really stupid, I'm on a foundation course having not done chemistry for 6 years since GCSE so struggling!)
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    (Original post by chemsitrynightmare)

    I don't suppose you're any good with dot and cross diagrams? It's the last question and I'm about to bang my head into a wall. It's for Sodium Nitrate I just can't work it out (sorry if i sound really stupid, I'm on a foundation course having not done chemistry for 6 years since GCSE so struggling!)

    Dude what have you been doing for the past 6 years? Its okay, no problem about the helping bit. I dont really have a problem helping you with the dot and cross but i dont know how i'll post it here. If you can tell me how to post pics i'll see if i can do it on 'ms paint' and attach it here.
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    The nitrate ion has 2x -1 charged oxygens, a +1 charged nitrogen and a neutral oxygen. Picture attached.

    Sodium will be +1, having lost an electron from it's outer shell.
    Attached Images
     
 
 
 
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