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10 June, 20:41

A chemistry graduate student is given 125. mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with = Ka*6.310-5. What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH = 4.63?

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  1. 10 June, 21:10
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    53.9 g

    Explanation:

    When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

    pH = pKa + log [A⁻]/[HA]

    where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

    We can calculate pKₐ from the given kₐ (pKₐ = - log Kₐ), and from there obtain the ratio [A⁻]/HA].

    Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

    So,

    4.63 = - log (6.3 x 10⁻⁵) + log [A⁻]/[HA] = - (-4.20) + log [A⁻]/[HA]

    ⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

    0.43 = log [A⁻]/[HA]

    taking antilogs to both sides of this equation:

    10^0.43 = [A⁻]/[HA] = 2.69

    [A⁻] / 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

    Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL (0.125 L) of a 2.69 M solution:

    (2.69 mol C6H5CO2⁻ / 1L) x 0.125 L = 0.34 mol C6H5CO2⁻

    The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 (160.21 g/mol):

    0.34 mol x 160.21 g/mol = 53.9 g
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