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7 February, 10:39

Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10-2, Ka2 = 6.5x10-8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.

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  1. 7 February, 10:58
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    pH = 7.1581

    Explanation:

    The equilibrium of NaHSO₃ with Na₂SO₃ is:

    HSO₃⁻ ⇄ SO₃²⁻ + H⁺

    Where K of equilibrium is the Ka2: 6.5x10⁻⁸

    HSO₃⁺ reacts with NaOH, thus:

    HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

    As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

    HSO₃⁻: 0.252 moles

    SO₃²⁻: 0.139 moles

    Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

    0.0500L ₓ (1mol / L) : 0.050 moles of NaOH.

    Thus, final moles of both compounds are:

    HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

    SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

    Using H-H equation for the HSO₃⁻ / / SO₃²⁻ buffer:

    pH = pka + log [SO₃²⁻] / [HSO₃⁻]

    Where pKa is - log Ka = 7.187

    Replacing:

    pH = 7.187 + log [0.189] / [0.202]

    pH = 7.1581
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