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24 August, 20:55

As the weak acid (HA) is titrated with a strong base (OH-), the neutralization reaction is:

HA + OH - → A - + H2O.

As the neutralization occurs, [HA] decreases and [A-] increases, allowing the solution to act as a buffer. Suppose 0.0015 mol HA and 0.0005 mol A - are present in 125 mL solution.

(a) What are the approximate concentrations of HA and A-?

(b) If the pKa of HA is 3.90, what is the pH of this solution?

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  1. 24 August, 23:48
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    a) [HA] = 0.012 M; [A-] = 4 E-3 M

    b) pH = 3.47

    Explanation:

    neutralization reaction:

    HA + OH - → A - + H2O

    ∴ mol HA = 0.0015 mol

    ∴ mol A - = 0.0005 mol

    ∴ Vsln = 125 mL = 0.125 L

    a) approx. concentration:

    ⇒ [HA] = 0.0015 mol / 0.125 L = 0.012 mol/L

    ⇒ [A-] = 0.0005 mol / 0.125 L = 4 E-3 mol/L

    b) pH = ?

    ∴ pKa = 3.90 = - Log Ka

    ⇒ Ka = 1.26 E-4 = ([H3O+] * (4 E-3 + [H3O+])) / (0.012 - [H3O+])

    ⇒ 1.512 E-6 - 1.26 E-4[H3O+] = 4 E-3[H3O+] + [H3O+]²

    ⇒ [H3O+]² + 4.126 E-3[H3O+] - 1.512 E-6 = 0

    ⇒ [H3O+] = 3.3866 E-4 M

    ⇒ pH = 3.4702
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