Weak acid dilution


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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 $\mathrm{pH} = 2.5$?
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> Consider volumes additive.
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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 $\mathrm{pH}$.




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.




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*Note that:*


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


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




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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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