Acid-base titration problem

Question

Consider the titration of 20.0 mL of 0.45 M HOCl (Ka = 1.3E-5) with 0.15 M Ca(OH)2. What is the pH of the solution after addition of a total of 20.0 mL of 0.15 M Ca(OH)2?

Even a basic run-through would help.... thanks

Answer 1

find moles:

0.0200 litres @ 0.45 mol / Litre = 0.00900 moles acid

0.0200 litres @ 0.15 mol/litre = 0.00300 moles Ca(OH)2

@ twice the moles of OH- per mole of Ca(OH)2 = 0.00600 moles OH-

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adding the 0.00600 moles of OH- destrroys an equal amount of acid, leaving only 0.00300 moles of acid left over,.... but it also produces 0.00600 moles of (OCl-) ion

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HOCl --> H+  &  OCl-

K = [H+]  [OCl-]  / [HOCl]

1.3e-5 = [H+]  [0.00600]  / [0.00300]

H+ = 6.5e-6

your answer: pH = 5.19