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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.

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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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