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24 August, 00:22

A beaker with 1.90*102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.60 mL of a 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

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  1. 24 August, 00:30
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    The change in pH is - 0.047

    Explanation:

    Let's first determine how much acetic acid and acetate we have in the buffer:

    pH = pKa + log (base/acid)

    5.000 = 4.740 + log (base/acid)

    0.260 = log (base/acid)

    100.240 = 10log (base/acid)

    base/acid = 1.73780

    1.90 x10^2 mL buffer (0.100 M) = 190 mmols of Acid + Base (A + B from now on)

    B/A = 1.7378

    B = 1.73780 (A)

    190 mmol = A + B

    190 mmol = A + 1.7378 (A)

    190 mmol = 2.7378A

    A = 69.39 mmol = acetic acid

    Amount of B: 190 mmol - 69.39 mmol = 120.60 mmol B = acetate

    Now that we have mili moles of A and B, we can see what remains after HCl is added:

    7.60 mL HCl (0.330 M) = 2.508 mmol HCl added

    Acetate + HCl - --> Acetic acid + H2O

    Before 120.60 2.508 69.39

    Change - 2.508 - 2.508 + 2.508

    Final 117.49 0 71.898

    We still have a buffer, so we can use pH = pKa + log (base/acid) again to find the pH:

    pH = 4.740 + log (117.49/71.898) = 4.9532

    ΔpH: 4.9532 - 5.000 = - 0.0467=-0.047
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