Friday
April 18, 2014

Homework Help: chemistry

Posted by Abigail on Tuesday, April 15, 2008 at 4:55pm.

Please let me know if my answer to the following question is correct.

Question:
25 mL of standardized 0.45 mol/L NaOH is titrated with 21 mL of 0.35 mol/L acetic acid. Calculate the pH of the solution.

Answer:
Add 25 mL and 21 mL to get 46 mL.
Add 0.45 mol/L and 0.35 mol/L to get 0.8 mol/L.
Divide 0.0046 by 0.8 to get 5.8 x 10^-3.
Since pH = -log[H^+], pH = -log[5.8 x 10^-3] gives the answer to this question, which is a pH of 2.2

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