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The following problem I have stated was not given to me as a homework. I have found this one in my textbook. I have solved it in my own way, but I'm not sure about the correctness of my procedure. I have posted to be sure about the procedure that I want to follow to solve such kind of problem.

A $5~\mathrm{ml}$ sample of $0.2~\mathrm{M}\ \ce{CH3COOH}$ is titrated with $0.2~\mathrm{M}\ \ce{NaOH}$. Calculate the $\mathrm{pH}$ after adding $25~\mathrm{ml}$ volume of base has been added. Where $K_\mathrm{a} = 1.8 \times 10^{-5}$

By applying $v_1s_1=v_2s_2$ formula, I got the concentration of weak acid and it is $0.0333~\mathrm{M}$. Again by applying same formula, for the total volume of $30~\mathrm{ml}$, I got the concentration of base and it is $0.1667~\mathrm{M}$.

Since the concentration of base is more than that of acid, $0.0333~\mathrm{M}$ base will neutralize the $0.0333~\mathrm{M}$ weak acid and the concentration of remaining base will be $0.1334\ \mathrm{M}$.

Then I used this concentration to evaluate the $\mathrm{pOH}$ of this solution and it was $0.875$. So $\mathrm{pH}$ will be $13.125$.

Do you think that my procedure to solve this problem is correct? I am confused. Because my textbook says that $\mathrm{pH}$ will be $8.87$. Has the author of this book made an error in this case? I cannot use Henderson-Hasselbalch equation in this case because it is not a buffer solution. If weak acid ($\ce{CH3COOH}$) and a strong salt of that weak acid ($\ce{CH3COONa}$) stay together in a solution, we can say it to be a buffer solution. But in the problem I stated above we can see that no weak acid exists after reaction takes place, and strong base ($\ce{NaOH}$) and strong salt ($\ce{CH3COONa}$) stay together. We also know that strong base and strong salt of that base cannot form buffer solution.

Now I want suggestion from the experts of this forum. Is my procedure correct? If it is not, which procedure should be followed to solve this problem?

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    $\begingroup$ Your reasoning looks about right (didn't check the math, though), and the problem itself looks broken. Who in their right mind would add five times the equivalent amount when titrating? And of course such a huge excess of a strong alkali is going to produce a pretty basic solution, nowhere near 8 or 9. $\endgroup$ – Ivan Neretin Jun 27 '16 at 9:58
  • $\begingroup$ @ Ivan Neretin, I cannot understand your explanation. Please make it easy, because English is my second language. Is the pH of this solution 8 or 9? How it is? Would you like to show it? $\endgroup$ – Nazmul Hassan Jun 27 '16 at 10:56
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    $\begingroup$ No, the pH is neither 8 nor 9. I think your answer is right. $\endgroup$ – Ivan Neretin Jun 27 '16 at 11:37
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    $\begingroup$ I know you wrote ‘this is not homework’ in the preamble of your question, but I still added the homework tag, since that also applies to self-study problems and the like, which is relevant to your question. $\endgroup$ – Jan Jun 27 '16 at 12:15
  • $\begingroup$ What is $v$ and what is $s$ in $v_1s_1 = v_2s_2$? $\endgroup$ – Jan Jun 27 '16 at 12:16
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The pH of the solution after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ is $13.12$. Your book or solution manual is wrong. The $\mathrm{pH} = 8.87$ at the equivalence point.

The Henderson Hasselbalch equation only applies in the buffer region - the part of the titration curve before the equivalence point. I have worked out the pH for a number of volumes of titrant and graphed the curve. The buffer region is labeled. You can see that past the equivalence point, the pH jumps to above 12 and stays there. You determined the pH after 25 mL correctly.

enter image description here

There is a point where the $\mathrm{pH=8.87}$ - the equivalence point (labeled with its pH). After $\mathrm{5\ mL}$ of $\ce{NaoH}$ solution is added, all acetic acid has been converted into the acetate ion. We can solve an equilibrium problem to determine the $\mathrm{pOH}$ and $\mathrm{pH}$ at that point. The following equation applies:

$$\ce{CH3CO2- + H2O <=> CH3CO2H + OH-}\ \ \mathrm{K_b=10^{-9.26}}$$

Since all acetic acid was converted into acetate, and the volume has doubled, the initial concentration of acetate is $\mathrm{0.1\ M}$. The initial volume of acetic acid is $\mathrm{0\ M}$ and the initial concentration of hydroxide is $\mathrm{10^{-7}\ M}\approx 0$.

$$\begin{array}{|c|c|c|c|}\hline \ & [\ce{CH3CO2-}] & [\ce{CH3CO2H}] & [\ce{OH-}] \\ \hline I & 0.1 & 0 & 0\\ C & -x & +x & +x \\ E & 0.1 -x & +x & +x\\ \hline \end{array}$$

If we plug in the values into the law of mass action, we can solve for $[\ce{OH}]$ and determine $\mathrm{pOH}$.

$$\begin{align} K_b &= \dfrac{[\ce{CH3CO2H}][\ce{OH-}]}{[\ce{CH3CO2-}]}\\ 10^{-9.26} &= \dfrac{x^2}{0.1-x}\end{align}$$

Solving for x gives

$$\begin{align} [\ce{OH-}] &=x=7.41\times 10^{-6}\\ \mathrm{pOH} &=5.13 \\ \mathrm{pH} &=8.87 \end{align}$$

Even if I had not made the assumption that the initial hydroxide concentration was zero, I would get the same $\mathrm{pH}$.

Your book has a typographical error. Either you were meant to determine the pH after $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ or the pH after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$. Both would yield a $\mathrm{pH} =8.87$ at the equivalence point.

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The NaOH will dominate the solution. Your math seems...correct...as far as it goes. If I were your instructor, and you turned your work in - as written, above, I'd mark it wrong. But perhaps you've not be taught about significant figures, or perhaps you're just keeping it short. But claim a pH to 5 significant figures when you're given only 1 to start with is what I'd consider an "epic fail". If your textbook claims 3 significant figures, then shame on it. The "correct" answer is 13. When I do similar problems, I find it most useful to convert concentrations to moles. This allows me to use my intuition to double check the reasonableness of the answer. (but in this case, since you didn't make an error, it wouldn't be necessary.) IOW, I'd convert the HOAc and the NaOH to moles, subtract and then convert the remaining NaOH back to Molarity and on to pH. Perhaps not necessary if you're proficient in working with units of concentration. It might be instructive to calculate the pH at the equivalence point...I'm wondering if that might be what the problem set-up meant...pKb being 9.3 for NaOAc...

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  • $\begingroup$ If you’re arguing sig figs, why not make it $1 \times 10^1$? $\endgroup$ – Jan Jun 27 '16 at 12:59
  • $\begingroup$ @Li Zhi, would you like to show how you would solve this same problem? It would be helpful if you showed me an example. I don't know how I will request you to show me your way which is more correct than the way I followed. By the way you have wtitten something about pKb. Please tell me why this is important when NaOH is a strong base. Is it necessary to use pKb when we are solving the problem stated above? $\endgroup$ – Nazmul Hassan Jun 27 '16 at 13:17

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