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Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. ?

The value of Ka for HClO is 2.9 × 10⁻⁸. An ice table would be extremely helpful. Thank you so much for any help. 

1 Answer

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  • 2 months ago
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    First determine number of moles of HClO and ClO- in the original  buffer:

    Moles HClO = 0.1000 L X 0.13 mol/L = 0.013 mol

    Moles ClO- = 0.1000 L X 0.37 mol/L = 0.037 mol

    Adding NaOH will quantitatively neutralize HClO forming more ClO-.

    After addition, 

    moles HClO = 0.010

    moles ClO- = 0.040

    Now, you can calculate the pH using the expression for Ka as:

    Ka = [H+][ClO-]/[HClO] = 2.9X10^-8

    [H+] (0.04)/0.010 = 2.9X10^-8

    [H+] = 7.25X10^-9

    pH = 8.14

    OR you can use the Henderson-Hasselbalch equation:

    pH = pKa + log [ClO-]/[HClO]

    pH = 7.54 + log (0.04/0.01) = 8.14

    In either case you can use moles directly rather than calculating molarities, since both numbers would be divided by the same total volume.

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