Posted by Chemistry Problem on Saturday, May 1, 2010 at 12:48pm.
moles acetic acid = M x L
moles acetate = M x L.
Add NaOH. mols added = M x L. NaOH reacts with acetic acid to produce more acetate at the expense of the acetic acid. Therefore, calulate moles NaOH added, subtract from acetic acid and add to acetate. Then M = moles/L, substitute base and acid into Henderson-Hasselbalch equation and solve for pH. I get something like 4.9 but that's an estimate and you need to confirm it with more accurate calculations.
Thank you Dr.Bob222!!!
A buffer is prepared by adding 300.0 mL of 2.0 M NaOH to 500.0 mL of 2.0 M CH3COOH. What is the pH of this buffer?
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