Monday, May 29, 2017

physical chemistry - Calculating pH in titration of monoprotic weak acid with strong base



25.0ml of 0.10M CHX3COOH is titrated with 0.10M NaOH. What is the pH after the addition of 30.0ml NaOH?



In broad strokes, as I understand it, the sodium hydroxide in this case would completely consume the acetic acid, leaving a solution of acetate ions, NaX+ and OHX.



What makes sense to me then is to figure that the pH is now dominated by the concentration of hydroxide ions, so the pH would be given by


pH=14(log([OHX])=11.96


where [OHX]=[(30.025.0)×104]mol(30.0+25.0)×103l


i.e., the remaining moles of hydroxide (0.0005) per liter of total volume (55ml)


but is this correct? Do I need to take into account the conjugate base’s contribution to deprotonation of water? If so, how is that done?


I’d like to be clever and use the lack of Ka or related constants in the problem to indicate that I’m on the right track, but that would be giving too much credit to my professor’s design of these questions (and the prior question in this series does give pKa for acetic acid)



Answer



Your calculation is right. Regarding your question to take into account the conjugate base's contribution to deprotonation of water, I'd say:


If you calculate the equilibrium constant of the neutralization reaction CHX3COOH+OHXCHX3COOX+HX2O K=KaKw=104.751014=1013.25


Now, if you write of reaction of acetate ion with water:CHX3COOX+HX2OCHX3COOH+OHX It's the opposite of the neutralization reaction. Its equilibrium constant is 1013.25.



You can see clearly that the neutralization reaction is the "predominant reaction". So, the pH is determined by this reaction as if it's the only one happening in the aqueous solution. So, you can neglect the reaction of of acetate ion with water.


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