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    A buffer solution was prepared by mixing 0.1 moldm3 ethanoic acid with sufficient sodium ethanoate so that the final concentration of sodium ethanoate was also 0.1 moldm3. The total vol. of the buffer was 100cm3. pka of ethanoic acid is 4.76.

    Calculate:
    -Mass of sodium ethanoate needed to make the buffer
    -ph of the buffer

    Thanks in advance
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    (Original post by cocoabeaneo)
    A buffer solution was prepared by mixing 0.1 moldm3 ethanoic acid with sufficient sodium ethanoate so that the final concentration of sodium ethanoate was also 0.1 moldm3. The total vol. of the buffer was 100cm3. pka of ethanoic acid is 4.76.

    Calculate:
    -Mass of sodium ethanoate needed to make the buffer
    -ph of the buffer

    Thanks in advance
    mass is simply the mass required to make 100ml of a 0.1 mol dm-3 solution.

    When acid concentration = salt concentration pKa = pH
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    (Original post by charco)
    mass is simply the mass required to make 100ml of a 0.1 mol dm-3 solution.

    When acid concentration = salt concentration pKa = pH
    Thanks for replying
    But how would you work out the ph of the buffer? The ph given is just for the acid
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    (Original post by cocoabeaneo)
    Thanks for replying
    But how would you work out the ph of the buffer? The ph given is just for the acid
    I repeat:

    When the concentration of acid and salt are the same, the pKa of the acid = pH of the solution.
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    (Original post by charco)
    I repeat:

    When the concentration of acid and salt are the same, the pKa of the acid = pH of the solution.
    Ah sorry I misread your reply. Okay that make sense, thanks!

    One more question: if the solution is diluted will the pH be the same again?
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    (Original post by cocoabeaneo)
    Ah sorry I misread your reply. Okay that make sense, thanks!

    One more question: if the solution is diluted will the pH be the same again?
    yes
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    Another Buffer soln question here
    Cacuate the p(OH) and the pH of an aqueous ammonia given that kb+1.8.10^-3 at 298k
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    (Original post by shady2.0)
    Another Buffer soln question here
    Cacuate the p(OH) and the pH of an aqueous ammonia given that kb+1.8.10^-3 at 298k
    .. show your working first.
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    (Original post by charco)
    .. show your working first.
    kb=[NH4][OH-]/[NH4OH]
    then
    -log[OH-]=-logkb+log[NH4+]/[NH4OH]
    since -log[OH-]=POH,POH+PH=14,then
    the equation will be
    14-PH=-log[1.8 times 10^-3]+log[NH4+]/0.1
    and that is where I got stuck
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    (Original post by shady2.0)

    Calcuate the p(OH) and the pH of an aqueous ammonia given that kb+1.8x10^-3 at 298k this should read 1.8 x 10[sup-5[/sup]

    kb=[NH4][OH-]/[NH4OH]

    then
    -log[OH-]=-logkb+log[NH4+]/[NH4OH] this line is confusing you. There is no need to go into the log form yet.

    since -log[OH-]=POH,POH+PH=14,then
    the equation will be
    14-PH=-log[1.8 times 10^-3]+log[NH4+]/0.1
    and that is where I got stuck
    kb=[NH4+][OH-]/[NH4OH]

    and you know that according to the stoichiometry of the reaction:

    NH3 + H2O --> NH4+ + OH-

    That [NH4+] = [OH-]

    hence

    kb * [NH4OH] =[OH-]2

    assuming that the ammonia solution is 0.1 mol dm-3 (you forgot to state it)

    and that very little interacts with the water ( a good approximation)

    then

    1.8 x 10-5 * 0.1 = [OH-]2

    [OH-] = √(1.8 x 10-6)

    [OH-] = 1.34 x 10-3

    pOH = 2.87

    pH = 11.13
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    (Original post by charco)
    kb=[NH4+][OH-]/[NH4OH]

    and you know that according to the stoichiometry of the reaction:

    NH3 + H2O --> NH4+ + OH-

    That [NH4+] = [OH-]

    hence

    kb * [NH4OH] =[OH-]2

    assuming that the ammonia solution is 0.1 mol dm-3 (you forgot to state it)

    and that very little interacts with the water ( a good approximation)

    then

    1.8 x 10-5 * 0.1 = [OH-]2

    [OH-] = √(1.8 x 10-6)

    [OH-] = 1.34 x 10-3

    pOH = 2.87

    pH = 11.13
    Thank you so much
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    (Original post by shady2.0)
    Thank you so much
    And sorry to bother again but when do u use the stoichiometry concept?and how do u use it?
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    (Original post by shady2.0)
    And sorry to bother again but when do u use the stoichiometry concept?and how do u use it?
    In any dilute weak acid or base situation.

    You can't use it with buffers.
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    (Original post by charco)
    In any dilute weak acid or base situation.

    You can't use it with buffers.
    ohkay gat it now thank you
 
 
 
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