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14 December, 12:54

A buffer solution based upon sulfurous acid (H2SO3) was created by treating 1.00 L of a 1.00 M sulfurous acid solution with NaOH until a pH of 1.135 was achieved (assuming no volume change). To this buffer 1.180 moles of NaOH were added (assume no volume change). What is the final pH of this solution

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  1. 14 December, 14:08
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    The final pH of this solution is 7.69

    Explanation:

    having the chemical reaction:

    H2SO3 = HSO3 - + H

    Ka = 1.5x10^-5

    pKa = 1.82

    H2SO3 + OH - = H2O + HSO3-

    1? 0

    1-x 0 + x

    the pH is equal to:

    pH = pKa + log ([HSO3-]/[H2SO3])

    1.135 = 1.82 + log (x / (1-x))

    Clearing x:

    x = 0.17

    Having the reaction:

    H2SO3 + OH - = H2O + HSO3-

    0.83 1.180 0.17

    0 0.83 1

    HSO3 - = SO3-2 + H+

    Ka2 = 1x10^-7

    pKa2 = 7

    HSO3 - + OH - = SO3-2 + H2O

    1 0.83 0

    0.17 0 0.83

    pH = pKa2 + log ([SO32-]/[HSO3-]) = 7 + log (0.83/0.17) = 7.69
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