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Homework Help: For Dr.Bob222 (chem work)

Posted by Gulzaman on Monday, August 27, 2007 at 11:17pm.

What is the pH of a solution of 120 ml 0.15M acetic acid to which we add 30mL 0.2M NaOH?

CAN YOU CHECK MY WORK PLEASE?

0.15 mol/L * 0.12 L = 0.018 mol acetic acid

0.2 mol/L * 0.03 L = 0.006 mol NaOH

0.018 - 0.006 = 0.012 mol of acetic acid in excess

pH = pKa + log (acid/base)

pH = 4.76 + log (0.006/0.018)

pH = 4.3

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