Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: brasarehot on January 25, 2012, 05:46:25 PM
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You titrate 1 L of 1 M PIPES by 10 N NaOH. Calculate the pH of the solution when 2ml of NaOH is added. ANSWER=5.06
Here's my work...but I got the wrong answer. What did I do wrong?
1Mx1L=1mols PIPES
10Mx(2/1000 L) = 0.02mols NaOH
(1-0.02)/(1L + 2/1000 L) = 0.978043912M remaining acid
pKa = [H+]^2 / [HA], 10^-6.76 = [H]^2/(0.978043912), [H] = 0.000412268
-log(0.000412268) = 3.384820823
So I'm getting pH=3.38 instead of 5.06....so what did I screw up? Thanks in advance