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24 December, 12:46

Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.

(a) pH of starting buffer

(b) pH after addition of 0.065 mol NaOH

(c) pH after further addition of 0.144 mol HCl

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  1. 24 December, 14:41
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    a) 4.73

    b) 4.78

    c) 4.66 (further addition) or 4.60 (starting from the original buffer solution)

    Explanation:

    Step 1: Data given

    volume of the buffer = 1.00 L

    Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

    pKa CH3COOH = 4.75

    Step 2: pH = pKa + log [CH3COONa]/[CH3COOH]

    pH = 4.75 + log (0.97/1.02)

    pH = 4.73

    (b) pH after addition of 0.065 mol NaOH

    Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A - in the same amount, increasing [A-] to 0.97 + 0.065 = 1.035

    pH = pKa + log[CH3COONa]/[CH3COOH]

    pH = 4.75 + log (1.035/0.955)

    pH = 4.78

    c) pH after further addition of 0.144 mol HCl

    The reverse will happen after the addition of HCl:

    [HA] = 0.955 + 0.144 = 1.099

    [A-] = 1.035 - 0.144 = 0.891

    pH = 4.75 + log (0.891/1.099)

    pH = 4.66

    If we add 0.144 mol of HCl to the original buffer we will get:

    [HA] = 1.02 + 0.144 = 1.164

    [A-] = 0.97 - 0.144 = 0.826

    pH = 4.75 + log (0.826/1.164)

    pH = 4.60
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