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    Dear all,
    I've been stuck on this question for ages.....is there any one that can help me.

    A solution is made by dissolving 7.5g NaOH, containing an inert impurity in water and making up to 250cm^3 of solution. If 20cm^3 of this solution is exactly neutralised by 13cm^3 of 1 mol dm-3 nitric acid, calculate the %age purity of the NaOH sample.

    I don't understand what I have to do....
    Do I find the number of moles in the nitric acid..(0.013mol) and then use that to find the mole in the reacted 20cm^3 of solution?
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    (Original post by Ayamma)
    Dear all,
    I've been stuck on this question for ages.....is there any one that can help me.

    A solution is made by dissolving 7.5g NaOH, containing an inert impurity in water and making up to 250cm^3 of solution. If 20cm^3 of this solution is exactly neutralised by 13cm^3 of 1 mol dm-3 nitric acid, calculate the %age purity of the NaOH sample.

    I don't understand what I have to do....
    Do I find the number of moles in the nitric acid..(0.013mol) and then use that to find the mole in the reacted 20cm^3 of solution?
    Yes..

    ...and then transfer moles to mass and find the actual mass of NaOH to find the % purity
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    so if the mole of HNO3 is 0.013 and one mole of HNO3 reacts with one mole of NaOH then the mole of the reacting NaOH is also 0.013 (i think).

    so does the mass of the reacting NaOH = 0.013 x (Mr of NaOH)
    23+16+1=40

    so 0.013 x 40 = 0.52g of NaOH reacted with the HNO3

    I managed to do all that but this is where I got stuck...im not sure what to do next..do i divide 0.5 by 7.5 and times by 100.

    I tried that but that doesnt give me the answer i am supposed to get....the answer I should get is 86.9%
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    I'd do it by working out the number of moles of NaOH that actually reacts with the HNO3.

    So:

    N = m/Mm

    This gives you 0.187 mol in 250cm3 solution. This in turn equals 0.748moldm-3. This is the concentration of the NaOH in the 20cm3 solution. Then use:

    N= C x V

    To work out the number of moles which reacts with the HNO3. Then take away the number of moles of HNO3 and you have the number of moles of the impurity.

    Hope that helped
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    Just to clarify what you are saying danhirons

    I work out the no of moles in NaOH in 250 solution 7.5/40= 0.1875 and find the concentration = 0.1875/0.25 =0.75 mol dm^3

    Did you mean that this should be the same concentration in the 20cm^3 solution? If so then the number of moles in the 20cm^3 solution is 0.75 x 0.02= 0.015. now the number of moles in the HNO3 solution is 0.013 and if this reacts exactly with the NaOH then 0.013 moles of NaOH solution reacts.....so do i do 0.015-0.013 which equals 0.002? if the impurity is 0.002 moles does that mean that it is 0.002 x 40= 0.008 grams?

    If that is correct do i minus 0.008 from 7.5 which equals 7.42?.... but if i use that and divide it by 7.5 i get 98.9%...so im still confused..where did i go wrong?
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    You have the number of moles of the impure NaOH, and the number of moles of HNO3. Then to calculate the impurity:

    ((the no. of moles difference between the two)/ the total number of moles of impure NaOH ) x 100 = percentage impurity - then take this away from 100

    If you don't understand why it's that I can explain
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    im sorry but i still don't understand....can u explain it to me with calculations rather than just words please coz dat makes it easier.
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    moles NH3 = conc x vol = 1 x 13/1000 = 0.013
    moles NaOH = 0.013 in 20cm3
    moles NaOH in 250 = 0.1625
    moles x mr = 0.1625 x 40 = 6.5g

    Purity = (6.5/7.5) x100 = 87% (86.7%)
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    thanks for your help...really grateful
 
 
 
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