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    Hi,

    Could someone check this for me?

    A titre of 23.45cm3 was obtained when ethanoic acid was titrated against 25.00cm3 of 0.100moldm-3 sodium hydroxide. Find the concentration of the ethanoic acid in mol/dm3 and g/dm3.

    n(NaOH) = c x v

    = 0.100 x (25 x 10-3)

    = 0.0025 moles therefore 0.0025 moles ethanoic acid.

    conc(CH3COOH) = n/v

    = 0.0025/0.002345

    = 0.107moldm-3

    conc in gdm-3 = conc in moldm-3 x Mr

    = 0.107 x 60.0

    = 6.40gdm-3

    Thanks!
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    looks right to me
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    well from the initial volumes given you'd expect the answer to be slightly more concentrated than the NaOH if its a 1:1 reaction, so would seem about right :dontknow:
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    (Original post by nexttime)
    well from the initial volumes given you'd expect the answer to be slightly more concentrated than the NaOH if its a 1:1 reaction, so would seem about right :dontknow:
    Yeah that's what I thought. Thanks both of you
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    i think that it is beneficial to put the molar ratio in it even if its 1:1 just in case, even though you have said it in words
 
 
 
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