# Problem 1

- Page ID
- 302704

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)**1. Calculate the pH of a 0.127 M solution of ascorbic acid (H _{2}asc).**

This is a new situation that we have not encountered before, since ascorbic acid (which is vitamin C by the way) has two dissociable hydrogen ions. We can find this out by looking in the table and seeing that two pK_{a} values (4.30 and 11.82) are provided. The relevant reactions needed to describe what happens in a solution of ascorbic acid in water are shown below.

\[\ce{H2asc + H2O \leftrightarrow Hasc- + H3O+}\qquad\qquad \mathrm{K_{a1}}\nonumber\]

\[\ce{Hasc- + H2O \leftrightarrow asc^2- + H3O+}\qquad\qquad \mathrm{K_{a2}}\nonumber\]

The problem we have is that there are two reactions that can cause production of H_{3}O^{+}. Remember that there is a small amount of H_{3}O^{+} in solution to begin with (10^{-7} M from the dissociation of water), but like in other problems of weak acids, we can assume that the H_{3}O^{+} produced by the dissociation of ascorbic acid will be much larger than the amount there from the dissociation of water.

The first step in understanding how to do this problem is to write an expression for H_{3}O^{+} in terms of the other species that are produced when the ascorbic acid dissociates. In other words, we ought to be able to write an expression that equates the concentration of H_{3}O^{+} to the concentrations of Hasc^{–} and asc^{2–}.

First consider the Hasc^{–} species. If we look at the first reaction above, it should be apparent that one H_{3}O^{+} occurs for each Hasc^{–} found in solution. If this was the only reaction that took place in solution, we could therefore write:

\[\mathrm{[H_3O^+] = [Hasc^-]}\nonumber\]

Next consider the asc^{2–} species. One way to think about this is to envision a situation in which all of the ascorbic acid dissociated into the asc^{2–} form. Under this situation, we could write the following reaction to describe this process.

\[\mathrm{H_2asc = asc^{2-} + 2 H_3O^+}\nonumber\]

The important thing to note here is that two H_{3}O^{+} occur for each asc^{2–} found in solution. Similarly, if only a small amount of asc^{2–} is found in solution, as would occur in this solution of ascorbic acid given the pK_{a} values, it would still be the case that two H_{3}O^{+} ions must occur for each asc^{2–} species found in solution.

\[\mathrm{[H_3O^+] = 2[asc^{2-}]}\nonumber\]

Note, there are two H_{3}O^{+} for every asc^{2–}. Substitute a “1” in for asc^{2–} in the equation above and you will get “2” for the H_{3}O^{+}.

We can combine these two into one equation that describes the total concentration of H_{3}O^{+} in a solution of ascorbic acid as follows:

\[\mathrm{[H_3O^+] = [Hasc^-] + 2[asc^{2-}]}\nonumber\]

What this means is that for each Hasc^{-} in solution we have one H_{3}O^{+}, and for each asc^{2–} in solution, we have two corresponding H_{3}O^{+} ions.

The next critical step is to consider the relative magnitudes of these two terms. If we examine the two pK_{a} values for the ascorbic acid reactions, and convert them to K_{a} values, note that the second reaction has a K_{a} value that is about 10^{7} times smaller than the first. This means that the extent of the second reaction is minimal compared to that of the first. In other words, the amount of H_{3}O^{+} formed by the second reaction is insignificant compared to how much H_{3}O^{+} is produced by the first. This means that:

\[\mathrm{[Hasc^-] >> 2[asc^{2-}]}\nonumber\]

and we can use the following approximation to describe this solution:

\[\mathrm{[H_3O^+] = [Hasc^-]}\nonumber\]

In other words, we only need to consider the first reaction to determine the pH of a solution of ascorbic acid. Even though this looked initially like it might be a complicated system, if we only need to consider the first reaction, solving this problem is identical to what we have done earlier when solving for the pH of a solution of a monoprotic acid. This raises the question of whether we can always simplify such a problem down to only one reaction. The answer depends in part on the relative magnitudes of the two pK_{a} values. If these two differed by two units (pK_{a1} = 3, pK_{a2} = 5), this represents a 100-fold difference in the extent of reaction and we can ignore the second reaction. If you were to examine the typical pK_{a} values in the table for polyprotic acids, you would notice that the relative values would almost always allow you to treat this comparable to a monoprotic system. Only in a few instances when the two pK_{a} values are almost identical would you need to treat this in a more complex manner by including both reactions in the problem.

What we will find in general with polyprotic acid/base systems is that we can almost always find one reaction that is significant and ignore the other reactions in the series. As we examine more problems of this variety, we will see how this will apply to other situations.

Now we can solve for the pH of this solution. Note, that there are two important assumptions. Because K_{a1} is small, the value of x is small compared to the starting concentration of H_{2}asc (0.0127 >> x). Also, since H_{2}asc is a weak acid, the amount of H_{3}O^{+} produced by dissociation of the acid will be much larger than the starting concentration of H_{3}O^{+} that exists due to the autoprotolysis of water (x >> 10^{-7})

\[\begin{align}

& &&\ce{H2asc}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{Hasc-} \hspace{25px} + &&\ce{H3O+}\nonumber\\

&\ce{Initial} &&0.127 &&0 &&10^{-7}\nonumber\\

&\ce{Equilibrium} &&0.127 - \ce{x} &&\ce{x} &&\ce{x}\nonumber\\

&\ce{Assumption} &&0.127 >> \ce{x} && &&\ce{x} >> 10^{-7}\nonumber\\

&\ce{Approximation} &&0.127 &&\ce{x} &&\ce{x}\nonumber

\end{align}\nonumber\]

\[\mathrm{K_{a1} = \dfrac{[Hasc^-][H_3O^+]}{[H_2asc]}=\dfrac{(x)(x)}{0.127}=5.0\times10^{-5}}\nonumber\]

\[\mathrm{x = [H_3O^+] = 2.52\times10^{-3}}\nonumber\]

\[\mathrm{pH = 2.6}\nonumber\]

Checking the approximation shows that this is valid.

\[\dfrac{2.52\times10^{-3}}{0.127}\times100 = 1.98\%\nonumber\]

As one final check that it was reasonable to ignore the second reaction, we can plug in the values of [Hasc^{–}] and [H_{3}O^{+}] calculated above into the K_{a2} expression. That leads to an interesting finding as seen below that [asc^{2–}] equals K_{a2}. Note how small this is, and therefore how little extra H_{3}O^{+} would come from the second reaction. Ignoring this reaction in the calculation of pH was a valid thing to do.

\[\mathrm{K_{a2}= \dfrac{[asc^{2-}][H_3O^+]}{[Hasc^-]} = \dfrac{[asc^{2-}](2.52\times10^{-3})}{(2.52\times10^{-3})} = 1.51\times10^{-12}}\nonumber\]

\[\mathrm{[asc^{2–}] = 1.51\times10^{-12}}\nonumber\]

## Contributors and Attributions

- Thomas Wenzel, Bates College (twenzel@bates.edu)
- Sourced from the Analytical Sciences Digital Library