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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.
- hcbiochemLv 72 months agoFavorite Answer
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-.
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.