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    the student adds 50. cm^3 of 0.250mol dm^-3 butanoic acid to 0.0500mol dm^-3 sodium hydroxide. A buffer solution forms.

    Calculate the pH of the buffer solution
    the Ka of butanoic acid is 1.51 x 10^-5 mol dm^-3

    I keep getting 4.12 but the answer is 4.22.... i will really appreciate it if anyone can help me out Thanks x
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    (Original post by Every Blue Moon)
    Ok well...

    Moles of Butanoic acid = 0.250 x 50 / 1000 = 0.0125
    Moles of NaOH = 0.05 x 50 / 1000 = 2.5 x 10^-3

    CH3CH2CH2COOH + NaOH --> CH3CH2CH2COONa + H2O

    so 0.0125 moles of butanoic acid react with 2.5 x 10^-3 moles of NaOH to form 2.5 x 10^-3 moles of CH3CH2CH2COONa and water (but water is not needed here) because butanoic acid is in excess and its a 1 to 1 reaction. all the CH3CH2CH2COONa dissociate into CH3CH2CH2COO- to give you 2.5 x 10^-3 moles of CH3CH2CH2COO-

    So 2.5 x 10^-3 moles of butanoic acid are used up to give: 0.0125 - 2.5x 10^-3 = 0.01 moles of Butanoic acid

    So from the Ka equation:

    [H+] = Ka [ CH3CH2CH2COOH] / [CH3CH2CH2COO-]
    = (1.51 x 10^-5) x (0.01) / (2.5 x 10^-3)
    =6.04 x 10^-5

    pH = 4.22 (2.d.p)
    Wow that is excellent thanks! but i don't get your explanation on why the moles change. how is butanoic acid in excess?
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    (Original post by Every Blue Moon)
    Haha no prob

    Basically u have 0.0125 moles of butanoic acid reacting with 2.5*10^-3 moles of NaOH, to give you your products. This is a 1 to 1 reaction, where you have 1 mole of everything in the chemical reaction. So if you use up 'X' number of moles of butanoic acid, u must also use 'X' moles of NaOH. Here you use all your NaOH because you have a lot more moles of butanoic acid (hence it is in excess) to react and the reaction goes to completion.

    So if 2.5*10^-3 moles of NaOH are used up, that leaves you with 0 moles of it after the reaction; this means 2.5*10^-3 moles of butanoic acid are also used up to give you (0.0125 - 2.5*10^-3) moles of butanoic acid after the reaction, which is 0.01 moles.

    You then use the rest in the Ka equation with 0.01 as ur [CH3CH2CH2COOH] value.
    haha perfect! You're a star thank you very much are you doing chemistry at uni or something?
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    (Original post by Every Blue Moon)
    Haha no prob

    Basically u have 0.0125 moles of butanoic acid reacting with 2.5*10^-3 moles of NaOH, to give you your products. This is a 1 to 1 reaction, where you have 1 mole of everything in the chemical reaction. So if you use up 'X' number of moles of butanoic acid, u must also use 'X' moles of NaOH. Here you use all your NaOH because you have a lot more moles of butanoic acid (hence it is in excess) to react and the reaction goes to completion.

    So if 2.5*10^-3 moles of NaOH are used up, that leaves you with 0 moles of it after the reaction; this means 2.5*10^-3 moles of butanoic acid are also used up to give you (0.0125 - 2.5*10^-3) moles of butanoic acid after the reaction, which is 0.01 moles.

    You then use the rest in the Ka equation with 0.01 as ur [CH3CH2CH2COOH] value.
    I find the explanation a bit weird though. You're subtracting the moles of NaOH from the moles of Butanoic acid, but how are you using the MOLES of butanoic acid to find the [H+]? Don't you need to use the concentration of butanoic acid, not the moles? (since Ka is made from the concentration of products over reactants) Baffleddd
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    (Original post by InadequateJusticex)
    I find the explanation a bit weird though. You're subtracting the moles of NaOH from the moles of Butanoic acid, but how are you using the MOLES of butanoic acid to find the [H+]? Don't you need to use the concentration of butanoic acid, not the moles? (since Ka is made from the concentration of products over reactants) Baffleddd
    You do use the concentrations....
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    (Original post by IShouldBeRevising_)
    You do use the concentrations....
    He subtracted the moles of NaOH from the moles of Butanoic acid and used that value...unless he converted the moles back to conc or I misread something. I haven't done this question myself but it was discussed in class
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    (Original post by InadequateJusticex)
    He subtracted the moles of NaOH from the moles of Butanoic acid and used that value...unless he converted the moles back to conc or I misread something. I haven't done this question myself but it was discussed in class
    No offence to the guy above but what he said was wrong you dont use the moles....

    You use the remaining moles of acid which he found correctly and the number of moles of the salt of the acid and convert these in to concentrations and bang them in to the equation.

    It took me 40 minutes to do this question and only after those 40mins and help from a great chemist @charco on here I figured it out..

    Yes this questions was hard here is a quote from the examiners report

    Only the very best candidates calculated the concentrations
    of butanoic acid and butanoate ions in the resulting buffer solution and obtained a correct pH
    value of 4.22 for all marks.
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    (Original post by IShouldBeRevising_)
    No offence to the guy above but what he said was wrong you dont use the moles....

    You use the remaining moles of acid which he found correctly and the number of moles of the salt of the acid and convert these in to concentrations and bang them in to the equation.

    It took me 40 minutes to do this question and only after those 40mins and help from a great chemist @charco on here I figured it out..

    Yes this questions was hard here is a quote from the examiners report

    Only the very best candidates calculated the concentrations
    of butanoic acid and butanoate ions in the resulting buffer solution and obtained a correct pH
    value of 4.22 for all marks.
    So he DID use the mole value. I know you use the concentration in the Ka equation hence why I was really confused. Well, at least now we kinda know what to expect if something difficult like this comes up...
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    (Original post by InadequateJusticex)
    So he DID use the mole value. I know you use the concentration in the Ka equation hence why I was really confused. Well, at least now we kinda know what to expect if something difficult like this comes up...
    Yeah hope so...
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    Thanks everyone this was very helpful (2 years later!)
 
 
 
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