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Please can someone explain this OCR Chemistry buffer question to me. Thanks! Watch

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    A chemist produces a buffer by mixing together the following:
    200cm^3 of 3.20 mol dm-3 HCOOH
    800cm^3 of 0.50 mol dm-3 NaOH

    Ka = 1.7x10^-4 mol dm-3

    Volume of buffer solution is 1dm^3

    Calculate the pH of the buffer.

    The answer for this question states:

    pH = 3.99

    Please could someone explain to me how this answer is reached. Thanks!
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    (Original post by _Piper_)
    A chemist produces a buffer by mixing together the following:
    200cm^3 of 3.20 mol dm-3 HCOOH
    800cm^3 of 0.50 mol dm-3 NaOH

    Ka = 1.7x10^-4 mol dm-3

    Volume of buffer solution is 1dm^3

    Calculate the pH of the buffer.

    The answer for this question states:

    pH = 3.99

    Please could someone explain to me how this answer is reached. Thanks!
    Before we can begin to plug in the numbers into our equation, we've got to figure out the moles of our HA and our A-

    HA is our acid: HCOOH
    A- is the salt formed: HCOONa

    Why? Well, if we just used the concentration of the acid and the base given, that's incorrect, because that won't be the concentration of the acid and the base if I put them in a buffer solution.
    The two are going to react, thereby changing the moles and thus the concentration.

    You know how to calculate moles?
    Spoiler:
    Show
     \mathrm{Moles} = \mathrm{Conc.} \times \mathrm{ Volume\ (dm^3)}

    In this case the moles of HA is going to be 0.64 mol.
    And the moles of NaOH is going to be 0.4 mol..
    You can now see which one is going to be excess.
    This is because they react in a 1:1 ratio, i.e. if we have 0.4 mol of NaOH it will react with 0.4 mol of HA.

    Thus the remaining moles of HA must be 0.64 - 0.4 = 0.24 mol.

    We know what the moles of A- is, it's 0.4 mol
    Spoiler:
    Show
     \mathrm{HCOOH\ +\ NaOH} \longrightarrow \mathrm{HCOONa\ + \ H_{2}O}
    Every 1 mole of NaOH forms 1 mole of A-, i.e. HCOONa

    Now from here we can calculate the pH.
    We can find the concentration of both our HA and A-, but it's redundant.

    Why? Because the volume is 1dm3 which means we're just dividing by 1, and that returns the same number.

    So... stick it into the equation.
    This should become a straightforward buffer calculation.
    Spoiler:
    Show
     \mathrm{i.e. \ [H^+]} = \dfrac{\mathrm{K_c} \times \mathrm{[HA]}}{\mathrm{[A^-]}}

    becomes...

     \mathrm{ [H^+]} = \dfrac{1.7 \times 10^{-4} \times 0.24}{0.4}

    Take the negative log of that number, and voila
 
 
 
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