Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: miaou5 on January 29, 2013, 06:29:43 AM
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Sorry for posting twice in one night, but my late-night studying session isn't getting me anywhere! This is re: the Henderson Hasselbalch equation
According to my gchem notes, dilution does NOT change pH (given the formula weak acid --> H+ + conjugate base), since [conjugate base]/[weak acid] ratio will stay the same (e.g., if you dilute the base to half its molarity, the acid conc. will also be halved, hence no change in the ratio). But I was just thinking--if you added increasing amounts of water, wouldn't the pH get closer to 7? So wouldn't dilution ultimately change the pH? Not sure why my prof never addressed this (among other things). Big thanks!
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HH equation is a crude approximation that works correctly at "quite normal" concentrations of buffer solutions. If you mean infinite dilution than you have practically pure water and pH should be close to 7.
Check wikipedia for Chariot equation
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Or check the nearly endless thread here: http://www.chemicalforums.com/index.php?topic=65325.0;topicseen
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Or this ooooold thread: http://www.chemicalforums.com/index.php?topic=5915.0
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This shouldn't require much reading up of those threads - as you know I'm sure, Ca=n/V (Ca is the analytical concentration of the acid, n is the analytical number of moles of the acid and V is the total volume of the solution). When you add H2O you increase V without increasing n. And [H+] is related to Ca (when Ca rises, so does [H+]). So, what happens if you dilute the solution? ...