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Given a solution with molarity $\pu{0.5 M}$ of a weak acid $\ce{HA}$ has $\mathrm{pH} = 1.670$. How much water in liters needs to be added to $\pu{0.150 L}$ of the first solution to get a $\mathrm{pH} = 2.5$?

Consider volumes additive.

My attempt was calculating the $K_\mathrm{a}$ of the acid using the first $\mathrm{pH}$ and then using it in the standard variation table used in these problems with the knowledge that the concentration of $\ce{H3O+}$ at the equilibrium are given by the last $\mathrm{pH}$.

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  • $\begingroup$ Can i add a photo? $\endgroup$
    – Mattiatore
    Dec 9, 2017 at 15:18
  • $\begingroup$ Absolutely (click the 6th icon on the left on the toolbox when you are in the edit mode), but the preferred method of sharing is text. Check out How can I format math/chemistry expressions here? for more info. I'd recommend to add picture first to prevent the question from closing, and then edit it correctly (I can help you with this if you add the picture in the next 15 min, I then have to leave) $\endgroup$
    – andselisk
    Dec 9, 2017 at 15:20
  • $\begingroup$ Thank you for the edit! I'd say that generally the approach looks OK, what is the problem here? Did you get the wrong answer? Regarding the formatting, > denotes a quotation, e.g. I used it to quote out the problem itself since it's been taken from a third party (a textbook, I presume). I gotta go; anyway, welcome to Chemistry.SE and good luck with the question! $\endgroup$
    – andselisk
    Dec 9, 2017 at 15:34
  • $\begingroup$ I'd like to be optimistic about it so I will tell that I got a different answer from the one that is given! Yes the problem is from a chemistry exercise book for engineers. I was having problems with the image uplaod because of the size limit. Thanks! $\endgroup$
    – Mattiatore
    Dec 9, 2017 at 15:39
  • $\begingroup$ @MattiaMariantoni - you don't really need to calculate the $K_a$. What is the simplified formula for the pH of a solution of a weak monoprotic acid? If you know that, you just need to write it once with $pH = 1.67$ and $C_{acid} = 0.5$, then again with $pH = 2$ and $C_{acid}$ as the unknown. Divide the first equation by the second, and you should get an expression in $C_{acid}$ alone. Solve that and you get the concentration that the acid needs to have in the final solution. The rest is stoichiometry. $\endgroup$ Dec 9, 2017 at 15:42

1 Answer 1

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The initial acid concentration is $\ce{C(HA)}=0.5\ \mathrm{M}$.

The weak dissociation of the acid follows:

$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}$$

We'll ignore the auto-dissociation of water as negligible. So only the dissociation of the acid plays on the $pH$:

$$K_a=\frac{[\ce{H3O+}]\times [\ce{A-}]}{[\ce{HA}]}$$

Stoichiometry now tells us:

$\ce{[HA]=C(HA)-10^{-pH}}$ and:

$\ce{[A-]=10^{-pH}}$ and:

$\ce{[H3O+]=10^{-pH}}$

Insert all this into the expression above to get the $K_a$.

Now with the second $pH$, insert the same expressions into the now known $K_a$, and calculate the new acid concentration, call it $\ce{C'(HA)}$.

Let the original volume be $V_1$ and the volume after dilution $V_2$, then because the number of moles of acid hasn't changed:

$$\ce{C(HA)}\times V_1=\ce{C'(HA)}\times V_2$$

Then $V_2-V_1$ is the volume that has to be added to increase the $pH$ as stated per the problem.


Note that:

$\ce{[HA]>>10^{-pH}}$

Which simplifies the calculation of $K_a$ because $\ce{[HA] \approx C(HA)}$.


It is also correct, if we use the simplified acid constant expression, that strictly speaking we don't need to calculate the $K_a$.

At $\ce{pH=1.670}$:

$$K_a=\frac{\Big(10^{-1.67}\Big)^2}{\ce{C(HA)}}$$

At $\ce{pH=2.5}$:

$$K_a=\frac{\Big(10^{-2.5}\Big)^2}{\ce{C'(HA)}}$$

Divide both expressions by each other, the $K_a$ drop out and find the ratio of the concentrations of acid. From there find the ratio and difference of volumes, as above.

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  • $\begingroup$ Same result of the book, although for the second pH we cannot neglect the $10^{-pH}$ term. $\endgroup$
    – Mattiatore
    Dec 9, 2017 at 16:25
  • $\begingroup$ Why not? It's even smaller than in the first case: 0.0032 mol/L. $\endgroup$
    – Gert
    Dec 9, 2017 at 16:27
  • $\begingroup$ But the acid concentration is also much smaller, correct me if I am wrong please. $\endgroup$
    – Mattiatore
    Dec 9, 2017 at 16:29
  • $\begingroup$ No, I think you're right. I haven't run any numbers. In any case you can use the correct expression in the second case and get the right result. $\endgroup$
    – Gert
    Dec 9, 2017 at 16:32

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