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What percent of a solution needs to be acetic acid for a particular $\mathrm{pH}$? I have 100% acetic acid, and want $1~\mathrm{L}$ of solution with a $\mathrm{pH}$ of 5. I've done some research, but I still do not understand how to solve this problem.

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closed as off-topic by Loong, Klaus-Dieter Warzecha, J. LS, M.A.R. ಠ_ಠ, Martin - マーチン Apr 19 '15 at 12:54

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  • $\begingroup$ The original question was fine as it is. But then someone edited and it seem's like that this is purely a homework question which it was not and was why I decided to answer. $\endgroup$ – Jun-Goo Kwak Apr 19 '15 at 15:09
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This is a dilution problem. So in this instance, you want to add deionized water to your acetic acid. However, the question is, how much water do you add?

It is necessary first to find the molarity of the acetic acid. On PubChem, we see that the molecular weight is $60.05196 \ \frac{\text{g}}{\text{mol}}$ and density is $1.0446 \ \frac{\text{g}}{\text{cm}^3}$ @ $25\ ^\circ \text{C}$.

If we have 1 L of acetic acid, then

$$ 1.0446 \ \frac{\text{g}}{\text{cm}^3} \cdot \frac{1000\ \text{cm}^3}{1\ \text{L}} \cdot \frac{1\ \text{mol}}{60.05196\ \text{g}} = 17.394\ \frac{ \text{mol}}{\text{L}}$$

We know that,

$$ \text{pH} = -\log{\ce{[H3O+]}} $$

which implies that the concentration of hydronium ions to make the solution with pH equal to 5 is, $\ce{[H3O+]} = 1 \cdot 10^{-5} \ \text{M}$. We make our RICE table,

$$\ce{HC2H3O2 + H2O -> C2H3O2- + H3O+}$$

\begin{array} {|c|c|c|c|c|} \hline \text{Initial conc.} & x \ \text{mol} & - & \text{0 mol} & \text{0 mol}\\ \hline \text{Change conc.} & -1 \cdot 10^{-5}\ \text{M} & - & 1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\ \hline \text{End conc.} & x - 1 \cdot 10^{-5}\ \text{M} & - & 1 \cdot 10^{-5}\ \text{M} & 1 \cdot 10^{-5}\ \text{M}\\ \hline \end{array}

$$K_\text{a} = \frac{[\ce{H3O+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(0.00001)(0.00001)}{(x - 0.00001)} = 1.75 \cdot 10^{-5} $$

$$\frac{(1 \cdot 10^{-10})}{(x - 0.00001)} = 0.0000175$$ $$1 \cdot 10^{-10} = 0.0000175x - 1.75 \cdot 10^{-10} $$ $$2.75 \cdot 10^{-10} = 0.0000175x$$ $$x = 1.6 \cdot 10^{-5}\ \text{M} = [\ce{CH3COOH}] $$

Thus, you go from a $17.394\ \text{M}$ solution of acetic acid, to a $0.000016\ \text{M}$, which is a factor of $1.09\cdot10^{6}$. You can then dilute the solution by the appropriate amount by placing the amount of acid you want in a volumetric flask, and then successively dilute it to the required pH.

Two assumptions,

  1. Deprotonation is small enough that the equilibrium concentration of the acid is approximately equal to the same as its initial concentration.
  2. Autoprotolysis of water does not significantly contribute to pH.
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    $\begingroup$ Very good answer, in particular the use (and demonstration) of the RICE table $\endgroup$ – user15489 Apr 19 '15 at 8:23
  • $\begingroup$ why use density? $\endgroup$ – RE60K Apr 19 '15 at 9:57
  • $\begingroup$ @ADG I used density to easily find the molarity of 100% acetic acid. If you look at the chain-link calculations above, you can see I get the molarity that I wanted. For reference, the pH is roughly 1 with purely 100% acetic acid which is interesting since acetic acid is a "weak acid". $\endgroup$ – Jun-Goo Kwak Apr 19 '15 at 15:07
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What percent of a solution needs to be acetic acid for a particular pH? I have 100% acetic acid, and want 1 L of solution with a pH of 5. I've done some research, but I still do not understand how to solve this problem.

You need to know that if you add water moles remian same so $\rm M_1V_1=M_2V_2$. You also need to know $\rm pH=-\log[H^+]$ form where you get concentration of $\rm H^+$. Now for weak acids the cubic equation used to determine pH gets some terms negligible hence we use $\rm [H^+]=\sqrt{K_aM_2}$ where $\rm K_a$ is very common and I remember it as $1.7\times10^{-5}\sim10^{-4.7}$.

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