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enthalpy of neutralisation-experimental HELP watch

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    You are provided with an unknown acid which has a relative formula mass of 135. The acid is a white chrystalline solid, and is very soluble in water. Preliminary experiment shows us that the acid is monoprotic. However you do not know if the acid is weak or strong.

    When the acid is neutralised by a solution of NaOH, the reaction is exothermic. Strong acid = -58 kJ/mol, weaker acids produce a smaller enthalpy change.

    Typical values:

    Acid | Enthalpy change
    **********************
    HCl |-57.9
    ----------------------------
    HNO3 |-57.6
    ----------------------------
    CH2ClCOOH |-53.4
    ----------------------------
    CH3COOH |-50.1
    -----------------------------
    HCN |-38.2

    So, the problem is, how would I use the enthalpy change of neutralisation to determine the relative strength of the acid.

    Standard NaOH concentrations available: 2M, 1M, 0.1M

    Decide on a suitable temperature rise for the reaction.

    I'm totally stumped on this.
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    me too. I'll come back to it when I'm more awake for another go.
    Is that AS?
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    (Original post by mik1a)
    me too. I'll come back to it when I'm more awake for another go.
    Is that AS?
    Yes, you got the exact same problem?
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    huh? no I don't have that problem, I was just saying I didn't know how to do that

    what's monoprotic? only one proton donor per molecule?
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    (Original post by mik1a)
    huh? no I don't have that problem, I was just saying I didn't know how to do that

    what's monoprotic? only one proton donor per molecule?
    Yes, so basically this means that when reacting with NaOH, it reacts perfectly to fomr 1 mol of water, from 1 mole acid + 1 mole NaOH
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    so if we know it's monoprotic how can we not tell if it's weak or strong?
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    (Original post by mik1a)
    so if we know it's monoprotic how can we not tell if it's weak or strong?
    The strength of any acid depends upon the extent to which it gives hydrogen ions in solution. If the concentration of H+ ions in a solution is high the acid is said to be strong. So the monoprotic hint is to tell you that the reaction between NaOH and the acid would reduce to water and a salt radily. So a less trickier calculation.
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    Enthalpy changes for strong acid/ strong base will all be roughly the same, as the same essential reaction is taking place in each

    H+(aq) + OH- (aq) ---> H2O


    Strong acids and alkalis will be pretty mch completely ionised in solution, so no energy is needed to break bonds, it's just the energy given out during the neutralisation, that's why they're the same.

    So I think you would work out how many moles of the acid you have, react with the NaOH and record the temperature rise, then work out the enthalpy change. If it's around 58kJ/mol then it's strong, less then it's weaker.

    For the temperature rise, I think the formula is enthalpy change = mass of reactants* heat capacity (usually just taken to be 4.2)*change in temp. Could you use this to work out a suitabel temperature rise?
    Are you actually supposed to carry out an experiment, or is this just about writing out the theory?


    Anyway, this is just off the top of my head.
    I hope you find it even slightly helpful.
 
 
 

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