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Problem 3

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    3. Calculate the pH of a solution prepared by adding 30 ml of 0.1 M hydrochloric acid to 60 ml of 0.080 M potassium malonate.

    The first key to solving this problem is to identify the nature of potassium malonate. If we look in the table we will find the species malonic acid (H2mal), a diprotic acid. Potassium malonate is therefore the species K2mal. When you add this to water, you would get two potassium cations and the malonate ion (mal2–). Since potassium is the cation of a strong base (potassium hydroxide – KOH), it does not react in any way and is a spectator ion.

    \[\ce{H2mal + H2O \leftrightarrow Hmal- + H3O+} \qquad \mathrm{K_{a1} = 1.40 \times 10^{-3}}\nonumber\]

    \[\ce{Hmal- + H2O \leftrightarrow mal^2- + H3O+}\qquad \mathrm{K_{a2} = 2.01 \times 10^{-6}}\nonumber\]

    So the malonate ion is a base. We have added hydrochloric acid, a strong acid to the solution. The hydrochloric acid will therefore undergo a neutralization reaction with the malonate ion to produce Hmal. If it turns out that all the mal2– gets used up in producing Hmal and there is still an excess of hydrochloric acid, the additional hydrochloric acid will then convert Hmal to H2mal. Remember, a strong acid will always lead to the neutralization of a base. Hmal can act as a base and accept another hydrogen ion to product H2mal.

    Next we need to calculate the moles of mal2– and hydrochloric acid that we have in solution.

    Moles of mal2–: \(\mathrm{\left(0.08\: \dfrac{mol}{L}\right)(0.060\: L) = 0.0048\: moles}\)

    Moles of HCl: \(\mathrm{\left(0.10\: \dfrac{mol}{L}\right)(0.030\: L) = 0.0030\: moles}\)

    The reaction that describes what will occur is as follows:

    & &&\ce{mal^2-}\hspace{25px} + &&\ce{H3O+} \hspace{25px}\leftrightarrow &&\ce{Hmal-} \nonumber\\
    &\ce{Initial} &&\mathrm{0.0048\: mol} &&\mathrm{0.0030\: mol} &&0\nonumber\\
    &\ce{Neutralization} &&\mathrm{0.0018\: mol} &&0 &&\mathrm{0.0030\: mol}\nonumber

    Note that there are appreciable amounts of both members of a conjugate pair, which constitutes a buffer. The only remaining question is whether we need to consider the other reaction that can occur for the Hmal.

    \[\ce{Hmal- + H2O \leftrightarrow H2mal + OH-}\nonumber\]

    It turns out that just like the case of ascorbic acid or sodium carbonate, there is so much distinction between the K values for the two reactions that we can ignore the second one and only consider the mal2– /Hmal reaction in determining the pH. Since we have a buffer, we can now use the appropriate Henderson-Hasselbalch expression for Ka2.

    \[\mathrm{pH=pK_{a2}+ \log\left( \dfrac{[mal^{2-}]}{[Hmal^-]}\right )=5.696+\log\left( \dfrac{0.0018}{0.0030}\right )=5.47}\nonumber\]

    If you go ahead and substitute the amount of Hmal and \(\ce{OH-}\) into the Ka1 expression, as shown below, you see that the amount of H2mal that forms is insignificant compared to the Hmal concentration and can be ignored.

    \[\mathrm{[Hmal^-] = (0.0030\: mol/0.090\: L) = 0.0333\: M}\nonumber\]

    \[\mathrm{[H_3O^+] = 3.39\times 10^{-6}}\nonumber\]


    \[\mathrm{[H_2mal] = 8.06\times10^{-5}}\nonumber\]

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    This page titled Problem 3 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Contributor.

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