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6 February, 21:23

A beaker with 1.80*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 8.50 mL of a 0.470 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. 6 February, 21:53
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    The pH will change 0.39

    Explanation:

    Step 1: Data given

    Volume of acetic acid : 180 mL

    pH = 5.00

    The total molarity of acid and conjugate base in this buffer is 0.100 M.

    We add 8.50 mL ( = 0.0085 L) of a 0.470 M HCl

    pKa of acetic acid = 4.74

    Step 2: Calculate initial concentrations

    x = concentration acid

    y = concentration conjugate base

    x + y = 0.100

    pH = pKa = log ([y]/[x])

    5.00 = 4.74 + log y/x

    0.26 = log y/x

    10^0.26 = 1.82 = y/x

    1.82 x = y

    x + 1.82 x = 0.100

    2.82 x = 0.100

    x = 0.0355 = concentration of acid

    y = 0.0645 = concentration of conjugate base

    Step 3: Calculate moles

    Moles = molarity * volume

    moles acid = 0.180 L * 0.0355 M = 0.00639 moles

    moles conjugate base = 0.0645 M * 0.180 L=0.01161 moles

    moles HCl = 8.50 * 10^-3 L * 0.470 M=0.003995 moles

    A - + H + = HA

    moles conjugate base = 0.01161 - 0.003995=0.007615 moles

    moles acid = 0.00639 + 0.003995=0.010385moles

    Step 4: Calculate total volume

    total volume = 180 + 8.50 = 188.5 mL = 0.1885 L

    Step 5: Calculate concentration

    concentration acid = 0.010385 moles / 0.1885 L = 0.0551 M

    concentration conjugate base = 0.007615 / 0.1885 = 0.0404 M

    Step 6: Calculate pH

    pH = 4.74 + log 0.0404/0.0551=4.61

    Step 7: Calculate pH change

    change pH = 5.00 - 4.61=0.39
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