Question

given 100ml of buffer that is .50M in HOCl and .40 M in NaOCl what is...

given 100ml of buffer that is .50M in HOCl and .40 M in NaOCl what is the pH after 10ml of 1.0M NaOH has been added?( Ka for HOCl=3.5*10^_8)
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Answer #1

when NaOH added

NaOH + HOCl --------------> NaOCl + H2O

moles of HOCl = 0.05

moles of NaOCl = 0.04

moles of NaOH added = 1.0 x 0.01 =0.01

final moles of HOCl = 0.05 - 0.01 = 0.04

final moles of NaOCl = 0.01 + 0.01 = 0.05

total volume = 100 + 10 = 110 mL = 0.11 L

Molarity of NaOCl = 0.05 / 0.11 = 0.45 M

molarity of HOCl = 0.04 / 0.11 = 0.36 M

pH = pKa + log [NaOCl] / [HOCl]

pKa = -log Ka = -log [3.5 x 10-8] = 7.45

pH = 7.75 + log [0.45] / [0.36]

pH = 7.85

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