Thursday, 10 September 2015

acid base - pH of ammonium acetate solution


I'm trying to calculate the pH of a 1M NHX4CHX3COO. I know that I'll have these reactions:


NHX4CHX3COONHX4X++CHX3COOX
NHX4X+NHX3+HX+
CHX3COOHCHX3COOX+HX+


I know the Kas of the last two, so I'm able to calculate the K of the first one (it's Ka1Ka2), which gives me these equations:



Ka1Ka2=(xy)(xz)1x
Ka1=y(y+z)xy
Ka2=z(y+z)xz


But I have only 2 independent equations (the first one is just the ratio of the second and third) and three variables so I'm unable to solve for [HX+], which is y+z...


What do I do?



Answer



Ok, I will follow the assumption proposed by @Zhe above (possible incorrect as he states, but please do not be confused by that).


In order to solve this problem we need two acidity constants: pka(ammonium ion) = 9.25 and pka(acetic acid) = 4.76.


First we state the proton balance (the amount of protons taken up have to be equal to the amount of protons given off in the system): Initially we have H2O and CH3COONH4.


Proton balance: [H3O+] + [CH3COOH] = [OH-] + [NH3]



At pH = 7, [H3O+] = [OH-] =10^-7 M At this pH the proton balance can be simplified as [CH3COOH] = [NH3]. The simplified proton balance will be true only at a pH that is exactly in the middle of the two pka-values. We get pH = (4.76 + 9.25)/2 = 7.005 ≈ 7 (only one significant figure is given, since you have stated the concentration as 1 M).


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