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### AuthorTopic: Calculating pH of acid/base titration?  (Read 1622 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### brasarehot

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##### Calculating pH of acid/base titration?
« on: January 25, 2012, 11:46:25 AM »

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
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