help needed from smarty parties..please

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jr8047

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Biochemistry test question:

A 10 mL solution of a .1 M monoproctic weak acid (pKa 6.0) is titrated with 10 mL of .11 M NaOH. The final pH after mixing is_____.

a. 13.2
b. 11.7
c. 10.2
d 6.0
e. 4.8





ps. d is the answer I put and it is wrong..why??? anyone know???
 
In this case the base has a higher concentration than the acid, so if you mix an equal amount of the two, the mixture should be weakly basic, not acidic...
 
Angle Jr. said:
In this case the base has a higher concentration than the acid, so if you mix an equal amount of the two, the mixture should be weakly basic, not acidic...


ok, I'm with you there, but that still leaves three other possible answers. Do you, by chance, know how to solve for the correct answer? Thanks in advance!
 
Im not 100% sure, but I think with titrations you use N1V1=N2V2, N being normality... ?
 
LazurasDC said:
with titrations isnt it normality1*volume1 = normality2*volume2


hmm...I don't follow you exactly. How would this lead me to the pH?
 
Since the .1M acid neutralizes .1M of the base, the [OH] of the mixture would be .01. Then the pOH of the mixture is log(1/[.01]) = log(100) = 2. So pH should be 14 - pOH = 12. The closest answer seems to be b.
 
Angle Jr. said:
Since the .1M acid neutralizes .1M of the base, the [OH] of the mixture would be .01. Then the pOH of the mixture is log(1/[.01]) = log(100) = 2. So pH should be 14 - pOH = 12. The closest answer seems to be b. I don't know if my logarithm calculation is right as I did it mentally (I got 30 on QR). Please verify it with your calculator.


Yes, you are awesome! That is the correct answer. Thank you so much; i have been trying to solve this with the henderson-hasselbalch equation and it just wasn't working out. You have cleared it up for me...thanks again!! 👍
 
hey, is it possible to use the H-H equations pH = pKa + log( conjugate base/weak acid or pOH = pKb + log(conjugate acid/weak base)
to solve this question? For some reason the pKa of weak acid is given. So, I thought it may be a trick question? 😎
 
After titration, we may think that we will have 0.01 M [OH-] left. But the final volume doubled up at the end so we have to divide the [ ] by 2, which is 0.005 M. The pOH is between 2 and 3, and since we are using logaritum, we will assume the actual pOH leans toward to 2 a little more.
 
Right, log(1/.005) = log200 = log(2*100) = log2 + log100 = 0.3 + 2 = 2.3
I would have been wrong if the choices were: b.12, c.11.7 😳

By the way it is useful to memorize that log2 = 0.3 and log3 = 0.48. Then you don't even need a calculator to do simple pH questions, as you can easily derive the following values (we weren't allowed to use a calculator in Japan).

log4 = 2log2 = 0.6
log5 = log(10/2) = log10-log2 = 1-0.3 = 0.7
log6 = log(2*3) = log2+log3 = 0.3+0.48 = 0.78
log7 = log√49≒(1/2)log50 = 0.5*(log5+log10) = 0.5*(0.7+1) = 0.85
log8 = 3log2 = 3*0.3 = 0.9
log9 = 2log3 = 2*0.48 = 0.96
 
Anyone know why HH equation doesn't apply here?
 
jr8047 said:
Anyone know why HH equation doesn't apply here?
At the titration point, we can assume that all the weak acid get neutralized, which is [weak acid] = 0. If we put 0 back to the HH equation, pH = pKa + log( conjugate base/weak acid), you will have infinite. On the other hand, HH equation mostly uses for finding out a buffer solution pH or pOH.

Don't worry too much on this question. The actural DAT question is not as tricky as this one is.
 
At the University of Tokyo, Dr. Noyori, our Nobel Prize-winning chemistry professor, taught us not to use the Henderson-Hasselbalch equation for questions of this kind. Instead, he taught us to use the Honda-Harakiri equation, p + q = 1, where p = - 2*[pKa] + log(1/[OH]) - 1, and q = pH.
In this case, p = - 2*6 + log[1/0.005] - 1 = - 12 + 2.3 – 1 = -10.7
Therefore, pH = q = 1 – p = 1 – (-10.7) = 11.7 👍


…seriously, as shinji said, you use the H-H equation when the solution contains relatively large quantities of both weak acid and its conjugate base, i.e. buffer solutions. For example, you would use the H-H equation when you want to calculate the pH of a buffer solution containing known concentrations of NH3 and NH4Cl.
 
hey y'all~

thanks for the help and tips; that clears it up. You are brilliant!
 
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