# For a solution that is 0.285 M HC_3H_5O_2 (propionic acid, K_a=1.3×10−5) and 0.0888 M HI, calculate the following?

Mar 17, 2017

We interrogate the equilibrium:

${\text{H"_3"CCH"_2"CO"_2"H"+"H"_2"O"rarr"H"_3"CCH"_2"CO"_2^(-)+"H"_3"O}}^{+}$

#### Explanation:

${\text{H"_3"CCH"_2"CO"_2"H"+"H"_2"O"rarr"H"_3"CCH"_2"CO"_2^(-)+"H"_3"O}}^{+}$

And this equilibrium is governed by a constant, ${K}_{a}$, where.......

${K}_{a} = \left(\left[\text{H"_3"CCH"_2"CO"_2^(-)]["H"_3"O"^+])/(["H"_3"CCH"_2"CO"_2"H}\right]\right) = 1.3 \times {10}^{-} 5$

But the $\left[{\text{H"_3"O}}^{+}\right]$ will be artificially high, because $\text{HI}$ has been added to solution, and its dissociation will be reasonably complete. Now we make the approximation that $x \cdot m o l \cdot {L}^{-} 1$ of $\text{propionic acid}$ dissociates, and thus we reformulate the equilibrium expression as follows:

${K}_{a} = \frac{\left(x\right) \left(0.0888 + x\right)}{0.285 - x} = 1.3 \times {10}^{-} 5$

And so we solve for $x$, by assuming that $0.0888 > x$, and $0.285 > x$. This approx. may not be justified:

${x}_{1} \cong \frac{1.3 \times {10}^{-} 5 \times 0.285}{0.0888} = 4.2 \times {10}^{-} 5$

Given this approx. we can recycle it back into the equilibrium expression, and see how $x$ evolves........

${x}_{1} \cong \frac{1.3 \times {10}^{-} 5 \times 0.285}{0.0888} = 4.2 \times {10}^{-} 5$

${x}_{2} = 4.17 \times {10}^{-} 5$. Since the values of $x$ do not change substantially, our approx., was correct and we can continue with this estimate.

And thus $\left[{\text{^(-)"O(O=)CCH"_2"CH}}_{3}\right] = 4.2 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

$\left[{\text{I}}^{-}\right] = 0.0888 \cdot m o l \cdot {L}^{-} 1$, because we have reasonably assumed that all the $\text{hydroiodic acid}$ has dissociated..........